Since the amount of substance stays equal, you can calculate the volume of your final solution $V_\mathrm{end}$, form your start concentration $c_\mathrm{start}(\ce{OH-})$ and your initial volume $V_\mathrm{start}$ and your target concentration $c_\mathrm{end}(\ce{OH-})$. Your working equation should be:
\begin{align} c_\mathrm{start}(\ce{OH-})\cdot V_\mathrm{start} &= c_\mathrm{end}(\ce{OH-})\cdot V_\mathrm{end} \\ V_\mathrm{end} &= \frac{c_\mathrm{start}(\ce{OH-})}{c_\mathrm{end}(\ce{OH-})}\cdot V_\mathrm{start} \end{align}
You can work out the start concentration for the initial pH and the target concentration from your final pH. Given (at $T=298.15~\mathrm{K}$, $p=1~\mathrm{atm}$):
\begin{align} \ce{pH}_\mathrm{start} &= 13 \\ c_\mathrm{start}(\ce{OH-}) &= 10^{-(14-\ce{pH})}~\mathrm{mol/L} = 0.1~\mathrm{mol/L} \\ V_\mathrm{start} &= 0.1~\mathrm{L} \\ \ce{pH}_\mathrm{end} &=11 \\ c_\mathrm{end}(\ce{OH-}) &=10^{-(14-\ce{pH})}~\mathrm{mol/L} = 0.001~\mathrm{mol/L} \end{align}
Hence
$$ V_\mathrm{end}= \frac{c_\mathrm{start}(\ce{OH-})}{c_\mathrm{end}(\ce{OH-})}\cdot V_\mathrm{start} = 10~\mathrm{L} $$
and therefore
$$ \Delta V = V_\mathrm{end}-V_\mathrm{start} = 9.9~\mathrm{L}= 9900~\mathrm{mL}. $$
Here's a quick way of solving this problem. You know that the pH of the solution is given by the concentration of hydronium cations, #"H"_3"O"^(+)#.
When you double the volume of the solution, you essentially halve the concentration of hydronium cations. This means that if you take #["H"_3"O"^(+)]_0# to be initial concentration of hydronium cations, you will have
#["H"_ 3"O"^(+)]_ "dil" = (["H"_ 3"O"^(+)]_0)/2 -># the concentration after the dilution
You know that
#color(purple)(bar(ul(|color(white)(a/a)color(black)("pH" = - log(["H"_3"O"^(+)]))color(white)(a/a)|)))#
You can thus say that after the solution is diluted, its pH will be
#"pH"_ "dil" = - log(["H"_ 3"O"^(+)]_"dil")#
This, in turn, is equivalent to
#"pH"_ "dil" = - log((["H"_ 3"O"^(+)]_0)/2)#
#"pH"_ "dil" = -[log(["H"_ 3"O"^(+)]_ 0) - log2]#
#"pH"_ "dil" = log2 - log(["H"_3"O"^(+)]_0)#
But since
#"pH"_ 0 = - log(["H"_ 3"O"^(+)]_0) = 2.600#
you will have
#"pH"_ "dil" = log2 + "pH"_ 0#
#color(green)(bar(ul(|color(white)(a/a)color(black)("pH" _ "dil" = 2.600 + log 2 = 2.901)color(white)(a/a)|)))#
The answer is rounded to three decimal places.
Finally, does this result make sense?
Diluting the solution means decreasing its concentration of hydronium cations, which in turn implies increasing the pH of the solution, i.e. making it less acidic.
Please do not block ads on this website. Imagine you have a bottle containing 100 mL of 0.100 mol L-1 HCl(aq) at 25°C and 1 atmosphere pressure. This is our stock solution and it has a pH of 1.00 What will happen to the pH of this solution if you add more water, that is, if you dilute the solution? First, let's calculate the moles of HCl(aq) in our stock solution using the relationship: n(HCl(aq)) = c(HCl(aq)) × V(HCl(aq)) n(HCl(aq)) = moles of HCl(aq) c(HCl(aq)) = concentration of HCl(aq) in mol L-1 = 0.100 mol L-1 V(HCl(aq)) = volume of solution in L = 100 mL = 100/1000 L = 0.100 L n(HCl(aq)) = c(HCl(aq)) × V(HCl(aq)) n(HCl(aq)) = 0.100 mol L-1 × 0.100 L n(HCl(aq)) = 0.0100 mol HCl(aq) is a strong acid, so it fully dissociates. This aqueous solution contains no HCl molecules, it only contains the ions H+(aq) and Cl-(aq) as shown in the balanced chemical equation for the acid dissociation given below: HCl(aq) → H+(aq) + Cl-(aq) Using the stoichiometric ratio (mole ratio) of HCl(aq) to H+(aq) (1:1) we can calculate the moles of H+(aq) in the solution, n(H+(aq)), as shown below: n(H+(aq)) = n(HCl(aq)) = 0.0100 mol This tells us that there are 0.0100 mol H+(aq) in our stock solution of 100 mL of 0.100 mol L-1 HCl(aq) with a pH = 1.00. Now imagine that we pour all 100 mL of this solution into a 1.00 L volumetric flask and fill it up to the mark with distilled water, that is, we dilute the original stock solution of HCl(aq). The same amount of H+(aq) is present, but the volume has increased by a factor of 10. And, since the concentration of H+(aq) has decreased, we expect the pH of the acidic solution to increase. Let's do the calculations and see if our predictions are correct...
No ads = no money for us = no free stuff for you!Effect on pH of Diluting an Aqueous Solution of a Strong Acid
What will happen to the pH of the solution after dilution? Prediction: concentration of H+(aq) will decrease by a factor of 10, that is, we predict the concentration of the diluted solution to be 1/10 × 0.100 mol L-1 = 0.0100 mol L-1.
Prediction: because pH is a logarithmic scale, a decrease in H+(aq) concentration by a factor of 10 will result in an increase in pH by 1 unit, that is, new pH = 1.00 + 1 = 2.00
We can calculate the final concentration of our diluted solution by using the relationship:
c = n ÷ V
c = concentration of H+(aq) in solution after dilution
n = moles of H+(aq) in solution after dilution = moles H+(aq) in stock solution = 0.0100 mol
V = volume of solution after dilution = 1 L
Substituting these values into the equation to find the concentration of H+:
| concentration (mol L-1) | = | moles of solute (mol) volume of solution (L) |
| c | = | n V |
| c(H+(aq)) | = | 0.0100 (mol) 1 (L) |
| = | 0.0100 mol L-1 |
We correctly predicted the concentration of the final solution to be 0.0100 mol L-1
Now we can calculate the pH of the final diluted solution using this concentration of H+(aq):
pH = -log10[H+(aq)] = -log10[0.0100] = 2.00
So we have also correctly predicted the pH of the hydrochloric acid solution after dilution!
Now imagine that you dilute our stock solution so much that the concentration of H+(aq) becomes incredibly small, say, c(H+(aq)) = 10-10 mol L-1.
What will be the pH of this new solution?
If we try to calculate the pH using c(H+(aq)) = 10-10 we will end up with a silly answer, as shown below:
pH = -log10[H+(aq)] = -log10[10-10] = 10
The answer is silly because, for an aqueous solution at 25°C and 1 atm a pH greater than 7 would be basic, and we know our solution must be acidic because we have added H+(aq) when the strong acid fully dissociated.
So what's the problem?
The problem is that we have ignored the contribution that the self-dissociation of water makes to the concentration of H+(aq) in our solutions.
That is, water also dissociates to a small extent, releasing H+(aq) into solution as shown below:
H2O(l) ⇋ H+(aq) + OH-(aq)
But the extent of this dissociation is very small (Kw = 10-14 at 25° and 1atm).
The self-dissociation of water contributes only 10-7 moles of H+ per litre of water ([H+(aq)] = 10-7 mol L-1)!
When the concentration of H+(aq) as a result of the strong acid dissociation is much larger than 10-7 mol L-1, we can ignore the contribution made by the water:
H+(aq) >> 10-7 mol L-1 ignore H+(aq) contributed by dissociation of water
However, when the concentration of H+(aq) due to the dissociation of strong acid is very small we can no longer ignore the H+(aq) contributed by the self-dissociation of water.
Consider our extremely diluted solution in which the dissoication of strong acid contributes 10-10 mol of H+(aq) per litre of solution.
The self-dissociation of water contributes significantly more to the hydrogen ion concentration, 10-7 mol H+(aq) per litre of solution.
Since [H+(from water)] >> [H+(from acid)] we ignore the contribution made by the acid!!
H+(aq) << 10-7 mol L-1 ignore H+(aq) contributed by dissociation of acid
So, for an aqueous solution of hydrochloric acid with a concentration of 10-10 mol L-1 the concentration of H+(aq) will be approximately 10-7 mol L-1 and the pH of this solution will be:
pH = -log10[H+(aq)] = -log10[10-7] = 7
As you continue to dilute a stock stock solution of strong acid, the pH of the resultant solutions increases until it reaches 7 (at 25°, 1 atm)
When an aqueous solution of acid is diluted with water:
- concentration of H+(aq) decreases
- pH of solution after dilution is greater than pH of solution before dilution
- pH of solution after dilution is closer to 7 (at 25°C, 1 atm) than before dilution
Do you know this?
Join AUS-e-TUTE!
Play the game now!
In the example above we started with 100 mL of a stock solution of HCl(aq) with [H+(aq)] = 0.100 mol L-1 and pH = 1.00 (25°C, 1atm).
We added water to this stock solution, we diluted it to 1.00 L, and found that the pH had increased to 2.00
Could we dilute the stock solution with water so that the resultant solution has a pH = 3.00 ? (a) What would be the volume of the diluted solution?
(b) How much water would we have to add to our stock solution to achieve this volume?
For the stock solution before dilution:
ci = concentration of H+(aq) in stock solution = [H+(aq)(stock))] = 0.100 mol L-1
Vi = volume of stock solution = V(stock) = 100 mL = 100/1000 = 0.100 L
For the final solution, or resultant solution, after dilution of the stock solution:
cf = H+(aq) concentration after dilution = 10-pH = 10-3.00 = 0.001 mol L-1
Vf = final volume of solution = ? L
We can use the dilution formula to calculate Vf:
Divide both sides of the equation by cf:
| ci × Vi cf | = | cf × Vf cf |
| ci × Vi cf | = | Vf |
Substitute in the known values to determine Vf, the volume of solution after dilution:
| ci × Vi cf | = | Vf |
| 0.100 × 0.100 0.001 | = | Vf |
| 10.00 L | = | Vf |
If we pour all 100 mL of 0.100 mol L-1 stock solution into a 10.00 L volumetric flask and fill up to the mark with distilled water we would achieve a resultant solution with a pH = 3.00
So, how much water has been added to the stock solution?
If we assume additivity of volumes, then,
final volume of solution = initial volume of solution + volume of water added
final volume = 10.00 L = 10.00 L × 1000 mL/L = 10, 000 mL
initial volume = 100 mL
volume of water added = ? mL
Substituting these values into the expression above:
final volume = initial volume + volume of water added
10,000 mL = 100 mL + volume of water added (mL)
Subtract 100 mL from both sides of the equation:
10,000 mL - 100 mL = 100 mL - 100 mL + volume of water added (mL)
9,900 mL = volume of water added (mL)
We need to add 9,900 mL of water to 100 mL of the stock solution in order to make a solution with a pH = 3.00
Do you understand this?
Join AUS-e-TUTE!
Take the test now!
Question 1: 25 mL of 0.0100 mol L-1 HCl(aq) is pipetted into a 250 mL volumetric flask and made up to the mark with water. Calculate the pH of the resultant dilute solution (25°C, 1 atm).
Solution:
(Based on the StoPGoPS approach to problem solving.)
- What is the question asking you to do?
Calculate the pH of the dilute solution
pH = ? - What data (information) have you been given in the question?
Extract the data from the question:
Acid is HCl(aq)
ci = [HCl(stock)] = 0.0100 mol L-1
Vi = volume of stock solution used = 25 mL = 25/1000 L = 0.0250 L
Vf = final volume of solution after dilution = 250 mL = 250/1000 L = 0.250 L
- What is the relationship between what you know and what you need to find out?
(a) HCl(aq) is a strong acid so [HCl(aq)] = [H+(aq)]
(b) cf = concentration of H+(aq) in final solution :
cf = (ci × Vi) ÷ Vf
(c) pH = -log10[H+(aq)] = -log10[cf]
- Solve the equations to find the pH of the solution
(a) HCl(aq) is a strong acid so [HCl(aq)(stock)] = [H+(aq)(stock)] = ci = 0.0100 mol L-1
(b) cf = concentration of H+(aq) in final solution :
cf = (ci × Vi) ÷ Vf
Substitute in the values and solve for cf
cf = (0.0100 mol L-1 × 0.0250 L) ÷ 0.250 L
cf = (0.000250 mol) ÷ 0.250 L
cf = 0.00100 mol L-1
(c) pH = -log10[H+(aq)] = -log10[cf] = -log10[0.00100] = 3.00
- Is your answer plausible?
pH of stock solution: pH = -log[H+] = -log[0.0100] = 2.00
Stock solution was diluted by a factor of 10 (25 mL to 250 mL) so final concentration = 1/10 × 0.0100 = 0.001 mol L-1
Because pH is logarithmic scale, the pH will increase by 1 unit, so the final pH = 2.00 + 1 = 3.00Since this agrees with the value we calculated above, we are confident that our answer is plausible.
- State your solution to the problem "pH of resultant solution":
pH = 3.00
Question 2: What volume of water must be added to 50.00 mL of 0.0200 mol L-1 HCl(aq) to obtain a solution with a pH of 3.7 (25°C, 1 atm)?
Solution:
(Based on the StoPGoPS approach to problem solving.)
- What is the question asking you to do?
Calculate the volume of water added to the stock solution
V(water) = ? mL - What data (information) have you been given in the question?
Extract the data from the question:
Acid is HCl(aq), a strong acid that fully dissociates in water
ci = concentration of stock HCl(aq) = 0.0200 mol L-1
Vi = initial volume of stock HCl(aq) before dilution = 50.00 mL = 50.00/1000 L = 0.0500 L
pH(final) = pH of solution after dilution with water = 3.7
- What is the relationship between what you know and what you need to find out?
(a) cf = [H+(aq)(after dilution)] = 10-pH(final)
(b) Vf = final volume of solution after dilution = (ci × Vi) ÷ cf
(c) Vf = Vi + V(water added)
(assume additivity of volumes) - Solve the equations to find the V(water added)
(a) cf = [H+(aq)(after dilution)] = 10-pH(final) = 10-3.7 = 0.000200 mol L-1
(b) Vf = final volume of solution after dilution = (ci × Vi) ÷ cf
Substitute the values into the equation and solve for Vf:
Vf = (ci × Vi) ÷ cf
Vf = (0.0200 mol L-1 × 0.0500 L) ÷ 0.000200 mol L-1
Vf = 0.00100 ÷ 0.000200 L-1 = 5.00 L = 5.00 L × 1000 mL/L = 5,000 mL
(c) Vf = Vi + V(water added)
(assume additivity of volumes)Substitute the values into the equation and solve for V(water added):
5,000 mL = 50.00 mL + V(water added)
Subtract 50.00 mL from both sides of the equation:
5,000 mL - 50.00 mL = 50.00 mL - 50.00 mL + V(water added)
4950 mL = V(water added)
- Is your answer plausible?
Work backwards: calculate the pH of the HCl(aq) solution assuming 4950 mL of water was added to 50.00 mL of 0.0200 mol L-1 HCl(aq): V = total volume of solution = 4950 + 50 = 5000 mL = 5000 ÷ 1000 = 5 L
n(H+) = n(HClstock) = c(HClstock) × V(HClstock) = 0.0200 × 50/1000 = 0.001 mol
[H+final solution] = n(H+) ÷ V = 0.001 ÷ 5 = 0.0002 mol L-1
pH(final solution) = -log10[H+] = -log10[0.0002] = 3.7Since this agrees with the value we were given in the question, we are confident that our answer is plausible.
- State your solution to the problem "volume of water added":
V(water added) = 4950 mL
Can you apply this?
Join AUS-e-TUTE!
Take the exam now!