# 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

### The molecular weight of NaCH_{3}COO is = 82.034 g/mole.

2.05 g NaCH_{3}COO = (2.05 g)/(82.034 g/mole) = 2.499 x 10^{-2} mole NaCH_{3}COO

The initial concentration of NaCH_{3}COO is (2.499 x 10^{-2} mole) / (0.100 L) = 0.2499 M. Since NaCH_{3}COO is an ionic compound containing an alkali metal ion (Na^{+}), it is soluble in aqueous solution, and the initial concentrations of Na^{+} and CH_{3}COO^{-} are both 0.2499 M. Since HCl is a strong acid, the initial concentrations of H_{3}O^{+} and Cl^{-} are both 0.100 M. Acetic acid, CH_{3}COOH, is a weak acid with ionization constant K_{a} = 1.85 x 10^{-5}

### which ionizes in water according to

CH_{3}COOH + H_{2}O <=> H_{3}O^{+} + CH_{3}COO^{-}

The addition of the sodium acetate causes this reaction to proceed to the left. Let x be the number of moles per liter of H_{3}O^{+} which combine with CH_{3}COO^{-} to form CH_{3}COOH. At equilibrium,

[CH_{3}COOH] = x

[CH_{3}COO^{-}] = 0.2499 - x

[H_{3}O^{+}] = 0.100 - x

The ionization constant of acetic acid is :

### K_{a} = [H_{3}O^{+}][CH_{3}COO^{-}] / [CH_{3}COOH] = (0.100 - x)(0.2499 - x) / x

Solving we get

x = 0.099988 M

[CH_{3}COOH] = x = 0.099988 M

[CH_{3}COO^{-}] = 0.2499 - x = 0.1499 M

[H_{3}O^{+}] = 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 H_{3}O^{+}, CH_{3}COOH, and CH_{3}COO^{-} are

[H_{3}O^{+}] = [(0.100 L)(1.234 x 10^{-5} M) + (0.006 L)(0.100 M)] / (0.106 L)

= 5.672 x 10^{-3} M

[CH_{3}COOH] = (0.100 L)(0.099988 M) / (0.106 L) = 9.433 x 10^{-2} M

[CH_{3}COO^{-}] = (0.100 L)(0.1499 M) / (0.106 L) = 0.1414 M

The additional HCl causes more of the CH_{3}COO^{-} to combine with H_{3}O^{+} to form CH_{3}COOH. Let x be the number of moles per liter of CH_{3}COO^{-} which combine with H_{3}O^{+}. At equilibrium, the new concentrations are

[H_{3}O^{+}] = 5.672 x 10^{-3} - x = A - x

[CH_{3}COOH] = 9.433 x 10^{-2} + x = B + x

[CH_{3}COO^{-}] = 0.1414 - x = C - x

where A = 5.672 x 10^{-3}, B = 9.433 x 10^{-2}, and C = 0.1414. We have

K_{a} = [H_{3}O^{+}][CH_{3}COO^{-}] / [CH_{3}COOH]

Solving we get:So x = 5.658 x 10

^{-3}M and the final concentration of H

_{3}O

^{+}is

### [H_{3}O^{+}] = 5.672 x 10^{-3} M - 5.658 x 10^{-3} M = 1.362 x 10^{-5} M

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