2.05g of sodium acetate was added to 100ml of 0.1 M HCl solution find the H+ ion concentration of the resulting solution if 6ml of 1M HCl is further added to it., What will be the new H+ concentration

Dear student

This is in 100 mL of solution. Now if 6.00 mL of 0.100 M HCl is added, we have a total of 106 mL of solution. The starting concentrations of H3O+, CH3COOH, and CH3COO- are  [H3O+] = [(0.100 L)(1.234 x 10-5 M) + (0.006 L)(0.100 M)] / (0.106 L)  = 5.672 x 10-3 M  [CH3COOH] = (0.100 L)(0.099988 M) / (0.106 L) = 9.433 x 10-2 M  [CH3COO-] = (0.100 L)(0.1499 M) / (0.106 L) = 0.1414 M  The additional HCl causes more of the CH3COO- to combine with H3O+ to form CH3COOH. Let x be the number of moles per liter of CH3COO- which combine with H3O+. At equilibrium, the new concentrations are  [H3O+] = 5.672 x 10-3 - x = A - x  [CH3COOH] = 9.433 x 10-2 + x = B + x  [CH3COO-] = 0.1414 - x = C - x  where A = 5.672 x 10-3, B = 9.433 x 10-2, and C = 0.1414. We have  Ka = [H3O+][CH3COO-] / [CH3COOH]

Solving we get:

So x = 5.658 x 10-3 M and the final concentration of H3O+ is

[H3O+] = 5.672 x 10-3 M - 5.658 x 10-3 M = 1.362 x 10-5 M

Regards

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