A solution of buffer contains 0.5 M of acetic acid and 0.5 Mof sodium acetate. p
ID: 688167 • Letter: A
Question
A solution of buffer contains 0.5 M of acetic acid and 0.5 Mof sodium acetate. pH change will be upon adding 1.0 ml of0.25 M HCl to 10 ml of this solution(Ka=1.75x10-5) (The answer is 0.044, can you explain what equation to use andhow to solve the problem, thanks) A solution of buffer contains 0.5 M of acetic acid and 0.5 Mof sodium acetate. pH change will be upon adding 1.0 ml of0.25 M HCl to 10 ml of this solution(Ka=1.75x10-5) (The answer is 0.044, can you explain what equation to use andhow to solve the problem, thanks) (The answer is 0.044, can you explain what equation to use andhow to solve the problem, thanks)Explanation / Answer
Formula: pH = pKa + log [base] / [acid] Data: [base]= 0.5 M [acid] = 0.5 M Ka= 1.75x10-5 pKa = 4.75 Upon substituting the data in theformula, pH = 4.756 + log ( 0.5 / 0.5 ) =4.756 Number of moles of sodium acetate = 0.5 M * 0.01L = 0.005 moles Number of moles of aceticacid = 0.5 M * 0.01L =0.005 moles Number of moles of HCl = 0.001 L * 0.25M = 0.00025 moles Chemical equation: CH3COO- (aq) +H+ (aq) ----------------> CH3COOH (aq) Before rxn (moles) 0.005 0.005 0.00025 After rxn (moles) 0.005 -0.00025 0.005 + 0.00025 = 0.00475 = 0.00525 Upon substituting the data in to the formmula, pH = 4.756 + log ( 0.00475 / 0.00525 ) = 4.712 pH = 4.756 - 4.712 =0.044Related Questions
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