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200 mL of a strong acid solution of pH 2...

200 mL of a strong acid solution of pH 2.0 is mixed with 800 mL of another acid solution of pH 3.0. The pH of the resultant solution is

A

`2.55`

B

`2.97`

C

`2.40`

D

`2.10`

Text Solution

Verified by Experts

The correct Answer is:
A
`[H^(+)] = 10^(-2) N " for Acid"_(1)`
`[H^(+)] = 10^(-3) " N for Acid"_(2)`
`[H^(+)]=N_(R)=(N_(1)V_(1)+N_(2)V_(2))/((V_(1)+V_(2)))`
`N_(R) = (2.8)/1000`
`[H^(+)] = N_(R) = 28 xx 10^(-4)` `pH = - log [H^(+)] `
`pH = - log 28 xx 10^(-4)`
` pH = 4 - log 28`
`pH = 4- [log 2 xx 2 xx 7 ] `
`pH = 4 - [ log 2 + log 2 + log 7 ] `
`pH = 4 - [ 0.3010 + 0.3010 + 0.845] `
`pH = 4 - 1.447`
`pH = 2.55`
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