A solution of buffer contains 0.1 M of acetic acid and0.1 M of sodium acetate. p

Question

A solution of buffer contains 0.1 M of acetic acid and0.1 M of sodium acetate. pH change will be upon adding 1.0 mlof 0.05 M HCl to 10 ml of this solution(Ka=1.75x10-5) (The anwer is decrease 0.444 unite, please explain whatequation to use to solve the problem and the steps to do it, alsocan you explain why it is decreased rather than increased,thanks) A solution of buffer contains 0.1 M of acetic acid and0.1 M of sodium acetate. pH change will be upon adding 1.0 mlof 0.05 M HCl to 10 ml of this solution(Ka=1.75x10-5) (The anwer is decrease 0.444 unite, please explain whatequation to use to solve the problem and the steps to do it, alsocan you explain why it is decreased rather than increased,thanks)

Explanation / Answer

When we mix CH3COOH (aq) and NaOH(aq) We know for a Buffer solution, pH = pKa + log{[salt] / [acid]}                                       CH3COO^- (aq) + H^+ (aq) .............> CH3COOH (aq) Before reaction CH3COO^- = 0.1mol /L * 10mL(10^-3L /1mL ) = 10^-3 mol Moles of H^+added = 0.05mol / L * ( 1.0mL (10^-3L /1mL)) = 5.0x10^-5mol After the reaction, moles of CH3COO^- decreases and the molesof CH3COOH increases. So after reaction, moles of CH3COO^- = 10^-3 mol -5.0x10^-5mol = 9.5x10^-4mol             Moles of CH3COOH = 10^-3mol + 5.0x10^-5mol = 1.05x10^-3 Hence, pH = - log (1.75x10^-5) + log {(9.5x10^-4) /(1.05x10^-3)}                  =4.757 - 0.043                 = 4.714 pH decreased by 0.043 units

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