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Explain the buffer action is a basic buffer containing equimolar ammonium hydroxide and ammonium chloride. (ii) Calculate the pH of a buffer solution consisting of 0.4 M CH_3COOH and 0.4 M CH_3COONa. What is the change is the pH after adding 0.01 mol of HCl to 500 ml of the above buffer solution. Assume that the addition of HCl causes negligible change in the volume. |
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Answer» Solution :Dissociation to buffer components `NH_4OH(aq) leftrightarrow NH_4^+ (aq)+OH^- (aq)` `NH_4Clto NH_4^+ +Cl^-` Addition of `OH^-` The added `H^+` ions are NEUTRALIZED by `NH_4OH` and there is no appreciable DECREASE in pH. `NH_4OH(aq)+H^+ to NH_4^+ (aq)+H_2O(l)` Addition of `NH_4^+ (aq)+OH^- (aq) to NH_4OH(aq)` The added `OH^-` ions react with `NH_4^+` to produce UNIONIZED `NH_4OH`. SInce `NH_4OH` is a weak BASE, there is no appreciable increase in PH. (ii) Ph of buffer `CH_3COOH (aq) leftrightarrow CH_3COO^-(aq)+H^+ (aq)` 0.4-a a a `CH_3COONa(aq) to CH_3COO^- (aq)+Na^+ (aq)` 0.4 0.4 0.4 `[H^+]=(K_s[CH_3COOH])/([CH_3COO^-])` `[CH_3COOH]=0.4-aapprox0.4` `[CH_3COO^-]=0.4+a approx0.4` `therefore[H^+]=(K_s(0.4))/((0.4))` `[H^+]=1.8 times10^-5` `thereforepH=-log(1.8 times10^-5)=4.74` ADDITON of 0.01 molHCl to 500 ml to buffer Added `[H^+]=(0.01mol)/(500mL)=(0.01mol)/(1//2L)` =0.02 M `CH_3COOH(aq) to CH_3COO^- + Na^+` 0.4-a a a `CH_3COON to CH_3COO^(-) +Na^+(aq)` 0.4 0.4 0.4 `CH_3COO^(-)+HCl to CH_3COOH+Cl^-` (0.02) 0.02 0.02 0.02 `therefore[CH_3COOH]=0.4-a+0.02=0.42-a approx 0.42` `[CH_3COO^-]=0.4+a-0.02=0.38+a approx 0.38` `[H^+]=((1.8 times10^-5)(0.42))/((0.38))` `[H^+]=1.99 times10^-5` `pH=-log(1.99 times10^-5)` `=5-log1.99` `=5-0.30` =4.70 |
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