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What should be the concentration of NaA, if its 50 mL solution of `0.10MNH_(3)` and `0.10MNH_(4)CI` without changing the pH by more than `1.0` unit ? Assume no change in volume. `(K_(a) for HA= 1.0xx10^(-5))` |
Answer» Let a be the concentration of NaA. After mixing, `M_(NaA)=(axx50)/(100)=(a)/(2)` and `M_(HA)=(0.2xx50)/(100)=0.1` Now `pH= -log K_(a)+log((["Salt"])/(["Acid"]))` `4= -1.0xx10^(-5)+log ((a//2)/(0.1))` `a=0.2M` |
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