A 23.0 ml solution of 0.105M CH3COOH is titrated with a 0.210 M KOH solution. Ca

Question

A 23.0 ml solution of 0.105M CH3COOH is titrated with a 0.210 M KOH solution. Calculate the pH after the following additions of KOH solution: (a) 10.0 ml (b) 11.5 ml (c) 15.0 ml (please answer with exact values and answer to this exact question with the exact numbers) Thank- you in advance!

Explanation / Answer

a) No of mmols of CH3COOH = 23*0.105 =2.415mmol No of mmols of KOH = 0.21*10 =2.1mmol So, no of mmol of CH3COOH left in the solution =2.415-2.1 =0.315 mmol For weak acid(acetic acid), CH3COOH =>CH3COO- + H+ At t=o c 0 0 At t c(1-x) cx cx Ka= (cx)^2 /c(1-x) x=(Ka/c)^1/2 = (1.74*10^-5 / 3.15*10^-4)^1/2 =0.235 pH = -log [H+] =-log [cx] =-log [3.15*10^-4*0.235] =-log [7.403*10^-5] = 4.1306 b)No of mmols of CH3COOH = 23*0.105 =2.415mmol No of mmols of KOH = 0.21*11.5 =2.415mmol So, all the acid and base in the solution gets neutralised pH=7 c)No of mmols of CH3COOH = 23*0.105 =2.415mmol No of mmols of KOH = 0.21*15 =3.15mmol So, no of mmol of KOH left in the solution =3.15-2.415 =0.735 mmol pH=14 - pOH =14 -(-log [OH-]) =14 +(log [OH-]) =14+ log(7.35*10^-4) =10.86628

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