A solution of buffer contains 0.5 M of acetic acid and 0.5 Mof sodium acetate. p
ID: 688290 • 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 please explain what equation touse and how to work the problem, simplicty would be greatly appreciated, thanks 0 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 please explain what equation touse and how to work the problem, simplicty would be greatly appreciated, thanks 0Explanation / 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 = 0.00525As volume is common for both the species, concentration can bereplaced by number of moles.
Upon substituting the data in to the formmula, pH = 4.756 + log ( 0.00475 / 0.00525 ) = 4.712 pH = 4.756 - 4.712 =0.044
Hope it will help you.
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