Question

Prepare 250mL of 0.5M sodium acetate buffer, pH of 4.7. I have 1M of acetic acid stock solution and 1M of sodium hydroxide stock solution. pKa of acetic acid= 4.7 I've gotten this far: pH=pKa + log ([base]/[acid]) 4.7 = 4.7 + log ([base]/[acid]) 0 = log ([base]/[acid]) 1 = ([base]/[acid]) 1 = 1/1 1=1, base and acid are equal quantities? In 250ml of solution 1M of base = (1mol/1) * 0.250L= 0.25mol 1M of acid = (1mol/1) * 0.250L= 0.25mol I donno, my brain is fried atm. it'd be great if you could add some input if you have any. If I fill out an ICE table: CH3COOH + OH- ---> CH3COO- + H20 I. 0.5 mol 0.25 mol ----> 0 0 C. -025mol +0.25mol ---> +0.25mol 0 E. +0.25mol 0mol ------> +0.25mol 0 Is this correct? Does this mean that i make a 250mL solution that is 125mL acetic acid, 62.5mL sodium hydroxide and 62.5ml water? :/

Answer 1

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