Question

A buffer at pH of 4.35 contains CH3COOH at a concentration of 0.60 M and CH3COONa at a concentration of 0.30 M. What will be the pH when 75.8 mL of 0.980 M NaOH is added to 450.4 mL of the buffer? The pKa of CH3COOH is 4.76 dlace of

Answer 1

As Base, NaOH is added to the buffer react with Acetic Acid and form Sodium ACetate. So amount of acid decreases and amount of Salt , Sodium acetate increases in the buffer solution.
The Buffer Solution pH = 4.35 Made of Autic acid (CH3COOH) and Socium acetate CCHE COONa) [city cool] = 0.60 (M); ICHCOONa] = 0-30(m) Given that - 75.8ml of 0.98(M) NaOH is added to 450-4 ml of the buffer. Amount of CH₃COOH in 450.4 ml buffer somtion 0.6 0 mol 450.40= 0.2703 mol х 1000 ml Amount of Chocoon Næ in 450. 4 me buffer solution 0.30 mol 450.4ml x 450.17 ml = 0.135 moli 1000ml Amount of Naoh added -75.8ml of 0.98/M) 75.8 ml x 0.981m) = 25 74.284 ml-m 0.074284 mul - 74.284 X10-3 Mole The added Naoth react with CH₃COOH and sm + H2O CH3COOH decreases and chicoona increases. Сн, ссон + NaOH - CH3COO NA! 0.2703 0.074 284 0.135 mol mol Before mol addition After (0-135 +0.074284) mol actition (0.2703- 0.074284) mc) = 0.2093 mol = 0.196 mol Now, total volume of solntion =(450.4+75:8) and 526.2 ml . After addition of Naoh, Concentration of Acetic acid, [CH₃ COOH] = 0.196 X1000 (M) 526.2 = 0.3725(M) Scanned with CamScanner

- 2 concentration of Cty Coona, 0 • 20963 526.2 Honda X1000 [citz Coona] 0.3977 (MS Henderson equation for Acidic buffer Solutim pH = pka + log [salt [Avid [CASCOona? pka (City cook) + logo [enty cooh = 4.76 + long 192029972) pH = 4.788 So, the final pH = 4.788 CSScanned with CamScanner