Question

A 1.0L buffer solution contains 0.100 molHC2H3O2 and 0.100 molNaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5. Part A Calculate the pH of the solution upon the addition of 0.015 mol of NaOH to the original buffer. Express your answer to two decimal places. pH = nothing Request Answer Part B Calculate the pH of the solution upon the addition of 10.0 mL of 1.00 MHCl to the original buffer. Express the your answer to two decimal places

Answer 1

solution : Initially the solution is an acid buffer. According to your question, (62 902] i.e.[CH3 CooH] = 0.100 mot (M) [Na CH3º2] .. [CHg coona ] = 0.100 moted (my Ka of CH3 C00H = 1:8 x 10-5 1 pka = -loglka) LOS : -1091.8810-5) de line 4.74 s .. Initially pH = pka + log [etty coo Na] TĽ CHz Cool] = 4.74 + log :100 [ 4.74 11 Part - A When 0.015 mol of Naoh is added to the original buffer, then then added oh ions would lower the acid concentration & would increase the salt coneen. tration by the same magnitude of 0.015 (M) Now, [Cty cool] = 0.1 –.015 = 0.085 (m) [CHz Coo Na] = 0.1 +0.015 = evo 0.115 (m) i [CHg coona] ? Now, pH = pka + log! [CHY COOH7 4.74 + log 0.115 = 4.8 7 0.085 185 Artis Wanita the pH of the solution upon the addition of 0.015 mol of NaOH to the original buffer Lins ca Scanned with CamScanner

part-B As According to your question, when looo me of 1.00 m Hel is added to the original buffer, then -the Htion in the concentration of added solution - 10.0 X 1.00 --- 1000 = 0.01(M) YA the added Htion increases the concentration of CHg cooh and lowers the concentration of salt by the same magnitude of 0.01 (m) NOW, [cHgcooH] = 0+1+0.01 = 0.11cm) [CH3 Coo Na] = 0.1-0.01 = 0.09(M). L . NOW pH = pka + loa o into = 4.74 + loa [CHg Coona] [снасоону. 0.09 = 4.65 the pH of the solution upon the addition of 10.0 ml of 1.00m Hl to the original buffer is 4.65 [Ans] one Scanned with CamScanner
The solution of your question is given in two parts.