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Calculate the pH of the following mixtur...

Calculate the `pH` of the following mixtures. Given `K_(a)` of `CH_(3)COOH=2xx10^(-5)` and `K_(b)` of `NH_(4)OH=2xx10^(-5)`
(a) `50mL` of `0.10 M NaOH+50mL` of `0.10 M HCl`.
(b) `50mL` of `0.10 M NaOH+50mL` of `0.10 M CH_(3)COOH`
(c )=`50mL`of `0.05M NaOH+50mL` of `0.10 M CH_(3)COOH`
(d) `50mL of 0.10 M NH_(4)OH+50mL of 0.05M HCl`
(e) `50mL of 0.10 M NH_(4)OH+50mL` of `0.10 M HCl`.
(f) `50mL` of `0.05 M NH_(4)OH+50mL` of `0.05 M CH_(3)COOH`.

Text Solution

Verified by Experts

`{:(,H^(+),+,OH^(-),rarr,H_(2)O),(t=0,5mmol,,5mmol,,-):}`
so `pH` of resulting solution `=7`.
b `{:(,CH_(3)COOH,+,OH^(-),rarr,CH_(3)COO^(-),+,H_(2)O),(t=0,2.5mmol,,5mmol,,,,-),(,-,,2.5,,2.5,,-):}`
`[OH^(-)]=(2.5)/(100)M=2.5xx10^(-2)M`
`pOH=3-log(2.5)=1.6`
`,.pH=12.4`
(C) `{:(,CH_(3)COOH,+,OH^(-),rarr,CH(3)COO^(-),+,H_(2)O),(t=0,5mmol,,2.5mmol,,,,),(,2.5,,-,,2.5,,):}`
`pH=pK_(a)+log(([CH_(3)COO^(-)])/([CH_(3)COOH]))=pK_(a)=4.7`
(d)`{:(,NH_(4)OH,+,H^(+),rarr,NH_(4)^(+),+,H_(2)O),(t=0,5mmol,,2.5mmol,,,,),(,2.5,,-,,2.5,,):}`
`pOH=pK_(b)+log(([NH_(4)^(+)])/([NH_(3)]))=4.7`
`:.pH=9.3`
(e)`{:(,NH_(4)OH,+,H^(+),rarr,NH_(4)^(+),+,H_(2)O),(t=0,5mmol,,5mmol,,0,,),(,0,,0,,5,,):}`
`pH=(1)/(2)[14-4.7-log0.05]`
`pH=5.3`.
(f)`{:(,NH_(4)OH,+,CH_(3)COOH,rarr,CH_(3)COONH_(4),+,H_(2)O),(t=0,2.5mmol,,2.5mmol,,,,),(,-,,-,,2.5,,):}`
`pH=7+(1)/(2)pK_(a)-(1)/(2)pK_(b)=7`.
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