Text Solution
4.749.268.784.63
Answer : C
Solution : `pK_(a)(NH_(4)^(+))=9.26` <br> `:. pK_(b)(NH_(3))=pK_(w)-pK_(a)(NH_(4)^(+))` <br> `=14-9.26=4.74` <br> On adding 15 m mol (50 mL ) `NH_(3)` the resulting solution contains 15 m mol ( in 100 mL ) of `NH_(4)` Cl and 5 m mol ( in 100 mL ) of `NH_(3)`. <br> `:. ["Salt"]=15xx10^(-2)M,["Base"]=5xx10^(-2)M` <br> `:. pOH =pK_(b)+log . (["Salt"])/(["Base"])` <br> `=4.74+log. (15xx10^(-2))/(5xx10^(-2))` <br> `=4.74 +log 3=4.74+0.48=5.22` <br> `pH =14-pOH =14-5.22=8.78`
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- Thread starter Shackleford
- Start date Aug 6, 2008
What is the pH of a solution prepared by mixing 50.0 mL of 0.300 M HCL with 450.0 mL of 0.400 M HIO3? HIO3 Ka = 1.6 x 10^-1
I first converted each concentration to its new molarity by figuring out the number of moles of each then dividing by the new volume of 0.5 L. I used the ICE method to determine the H+ concentration. I got a pH of 0.714.Homework Equations
ICE
The Attempt at a Solution
Answers and Replies
0.714 looks OK
&
2
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