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20 mL of 0.1 M weak acid HA(K(a)=10^(-5)...

20 mL of 0.1 M weak acid `HA(K_(a)=10^(-5))` is mixed with solution of 10 mL of 0.3 M HCl and 10 mL. of 0.1 M NaOH. Find the value of `[A^(-)]`//([HA]+[A^(-)])` in the resulting solution :

A

`2xx10^(-4)`

B

`2xx10^(-5)`

C

`2xx10^(-3)`

D

0.05

Text Solution

Verified by Experts

The correct Answer is:
A
`[H^(+)]=(10xx0.3 – 10xx0.1)/(10+10+20)=0.05`
`[HA]=(20xx0.1)/(20+20) = 0.05`
`underset(0.05-x)(HA) to underset(0.05+x)(A^(+)) + underset(x)(A^(-))`
Due to common ion effect neglect x w.r.t. 0.05
`K_(a)= ((0.05 + x)x)/((0.05 –x))=x`
`therefore x=10^(-5) therefore [A^(-)]/([HA] + [A^(-)]) = x/(x+0.05)=2xx10^(-4)`
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