Question #191808

A one liter buffer solution was made of ammonia (0.33 M)-ammonium chloride (0.33 M) at equilibrium. A student was asked to add 0.05 mol HCl to that solution. However, he mistakenly added 0.07 mol NaOH to that solution. Assume no change in the volume of the buffer. a) What type of buffer solution it was. Provide its mechanism. b) Calculate the change of pH of the buffer after the mistake of the student c) What would be the change of pH if the student adds 1000 mL water to that buffer solution?


1
Expert's answer
2021-05-12T06:27:36-0400

Ans:-

If he mistakenly added 0.070.07 mol NaOH to that solution then its concentration [OH]=0.07M[OH^-]=0.07M because volume kept 1L1L

NH4++OHH2O+NH3NH_4^++OH^-\to H_2O+NH_3

[NH3]=0.33,[NH4+]=0.33[NH_3]=0.33 , [NH_4^+]=0.33

after adding 0.070.07 mol NaOH the concentration is [NH3]=0.40[NH_3]=0.40 , [NH4+]=0.26[NH_4^+]=0.26

pH=pKa+log[NH3][NH4+]\Rightarrow pH=pK_a +log\dfrac{[NH_3]}{[NH_4^+]}

pH=9.25+log[0.40][0.26]=9.98\Rightarrow pH=9.25 +log\dfrac{[0.40]}{[0.26]}=9.98

(b) Resulting buffer is slightly basic in nature because its pH will be greater than 77


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