What will be the pOH of a solution formed by mixing 40 cm³ of 0.1 M HCI with 10 cm³ of 0.45 m Naoh?

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Text Solution

`12.0``10.0``8.0``6.0`

Answer : A

Solution : `40 ml` of `0.1 M HCl` <br> `(0.1)/(1000)xx40=0.004` moles of `HCl` <br> Similarly, `10ml` of `0.45 M NaOH` <br> ` =(0.45)/(1000)xx10=0.0045` moles of `NaOH` <br> Thus, `NaOH` left unneutralised `=0.0005 ` moles in a volume of `50 ml` <br> Concentration of `NaOH=(1000xx0.0005)/(50)=0.01` <br> `=10^(-2) M` <br> Therefore, `[OH^(-)]=10^(-2)M` or `[H^(+)]` <br> ` =10^(-12) M` or `pH=12.0`

Text Solution

`10``8``5``12`

Answer : D

Solution : No of m eq of `H^(+)` ions `= 40 xx 0.1 = 4` <br> No of m.e of `OH` ions `= 10 xx 0.45 = 4.5` <br> `[O^(-)H] = (4.5 - 4)/(50) = (0.5)/(50) = 0.01` <br> `p^(OH) = -10 g 10^(-2), p^(OH) = 2` <br> `= p^(H) = 12`