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

The molecular weight of NaCH3COO is = 82.034 g/mole. 

2.05 g NaCH3COO = (2.05 g)/(82.034 g/mole)  = 2.499 x 10-2 mole NaCH3COO 

The initial concentration of NaCH3COO is (2.499 x 10-2 mole) / (0.100 L) = 0.2499 M. Since NaCH3COO is an ionic compound containing an alkali metal ion (Na+), it is soluble in aqueous solution, and the initial concentrations of Na+ and CH3COO- are both 0.2499 M. Since HCl is a strong acid, the initial concentrations of H3O+ and Cl- are both 0.100 M. Acetic acid, CH3COOH, is a weak acid with ionization constant Ka = 1.85 x 10-5 

which ionizes in water according to 

CH3COOH + H2O <=> H3O+ + CH3COO- 

The addition of the sodium acetate causes this reaction to proceed to the left. Let x be the number of moles per liter of H3O+ which combine with CH3COO- to form CH3COOH. At equilibrium, 

[CH3COOH] = x 
[CH3COO-] = 0.2499 - x 
[H3O+] = 0.100 - x 

The ionization constant of acetic acid is :

Ka = [H3O+][CH3COO-] / [CH3COOH] = (0.100 - x)(0.2499 - x) / x 
Solving we get
x = 0.099988 M 
[CH3COOH] = x = 0.099988 M 
[CH3COO-] = 0.2499 - x = 0.1499 M 
[H3O+] = 0.100 - x = 1.234 x 10-5 M 

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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