concentration of acetic acid= .2945 M A pH 4.80 buffer solution is prepared by a

Question

concentration of acetic acid= .2945 M

A pH 4.80 buffer solution is prepared by adding solid sodium acetate trihydrate NaC_2H_3O_2 3H_2O to 50.0 mL of the acetic acid solution that was standardized in problem 1. Use Equation (22-3) with the acetic acid concentration calculated in problem 1, the K_a for acetic acid (K_a = 1.8 Times 10^-5) and the concentration of H^+ calculated in problem 2 to calculate (a) the concentration of acetate in the buffer solution and (b) the mass of sodium acetate trihydrate required to prepare the solution with that concentration. Assume the addition of the solid does not change the solution volume. Remember that the water in the formula of a hydrate must be included in calculating molar mass of a hydrate.

Explanation / Answer

[H+] = 1.588 x 10^ -5 M

[CH3COOH] = 0.2945 M

we have equation

CH3COOH (aq) <---> CH3COO-(aq) + H +(aq)

Ka = [CH3COO-] [H+] / [CH3COOH]

1.8 x 10^ -5 = ( 1.585x10^-5) x [ CH3COO-] / ( 0.2945)

[CH3COO-] = 0.3344 M   is concnetrate of acetate in buffer

b) moles of CH3COO- = M x V = 0.3344 x ( 50/1000) = 0.01672            ( volume = 50 ml = 50/1000 L)

now acetate is given by sodium acetate trihydrate

hence moles of sodium acetate trihydrate = 0.01672

mass of sodium acetate trihydrate = moles x molar mass of sodium acetate trihydrate

                                         = 0.01672 x 136.079

                                        = 2.275 g

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