A solution with a volume of 1.00 L is 0.450 M in CH3COOH(aq) and 0.350 M in CH3C

Question

A solution with a volume of 1.00 L is 0.450 M in CH3COOH(aq) and 0.350 M in CH3COONa(aq). What will the pH be after 0.0400 mol of HCl is added to the solution? Ka for CH3COOH(aq) = 1.80 x 10-5
A solution with a volume of 1.00 L is 0.450 M in CH3COOH(aq) and 0.350 M in CH3COONa(aq). What will the pH be after 0.0400 mol of HCl is added to the solution? Ka for CH3COOH(aq) = 1.80 x 10-5
A solution with a volume of 1.00 L is 0.450 M in CH3COOH(aq) and 0.350 M in CH3COONa(aq). What will the pH be after 0.0400 mol of HCl is added to the solution? Ka for CH3COOH(aq) = 1.80 x 10-5

Explanation / Answer

mol of HCl added = 0.04 mol

CH3COONa will react with H+ to form CH3COOH

Before Reaction:

mol of CH3COONa = 0.35 M *1.0 L

mol of CH3COONa = 0.35 mol

mol of CH3COOH = 0.45 M *1.0 L

mol of CH3COOH = 0.45 mol

after reaction,

mol of CH3COONa = mol present initially - mol added

mol of CH3COONa = (0.35 - 0.04) mol

mol of CH3COONa = 0.31 mol

mol of CH3COOH = mol present initially + mol added

mol of CH3COOH = (0.45 + 0.04) mol

mol of CH3COOH = 0.49 mol

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.745+ log {0.31/0.49}

= 4.546

Answer: 4.55

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