# Solubility Question Need Help QUICK?

How would you prepare 1.0 L Buffer having a pH of 5.00 using of KC2H3O2 solid and 2.0M HC2H3O2 solution (Ka = 1.8x10^-5)

What would be the pH of a .50L Buffer after adding
a) .01 mole of KOH to buffer to above solution
b) .25 mol of HCL to buffer to above solution

Please help!! the test is tomorrow and I'm up late studying, I don't really know how to do these

Answer:
About the first question we can use the equation
pH = pK + log [CH3COO-] / [CH3COOH]
pK = - log K = - log 1.8 x 10^-5 = 4.74
5.00 = 4.74 + log [CH3COO-] / 2.0
0.26 = log [CH3COO-] / 2.0
10^0.26 = [CH3COO-] / 2.0
1.82 = [CH3COO-] / 2.0
[CH3COO-] = 3.64 M

To prepare 1.0 L of this buffer we need 3.64 moles of CH3COOK ( molar mass = 98.1 g/mol)
3.64 mol x 98.1 g/mol = 357 g of CH3COOK

Moles of CH3COOH in 0.50 L = 2.0 x 0.50 = 1.0
moles of CH3COOK in 0.50 L = 3.64 x 0.50 = 1.82

Moles KOH added = 0.1
The effect of the added OH- is to convert CH3COOH in CH3COO- via the net reaction :
CH3COOH + OH- >> CH3COO- + H2O

we get 1.0 - 0.1 = 0.9 moles CH3COOH
and 1.82 + 0.1 = 1.92 moles CH3COO-
[CH3COOH] = 0.9 / 0.5 L = 1.8 M
[H3COO-] = 1.92 / 0.5 = 3.84 M

pH = 4.74 + log 3.84 /1.8 = 5.07

moles CH3COOH = 1.0
moles CH3COO- = 1.82
The effect of the added 0.25 mole H+ would be to decrease the moles of CH3COO- by 0.25 and increase the moles of CH3COOH by 0.25
Moles CH3COOH = 1.0 + 0.25 = 1.25
moles CH3COO- = 1.82 - 0.25 = 1.57
[CH3COOH] = 1.25 / 0.5 = 2.5 M
[CH3COO-] = 1.57 / 0.5 = 3.14 M

pH = 4.74 + log 3.14 / 2.5 = 4.83

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