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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 )=50mLof 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. |
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Answer» Solution :`{:(,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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