Solubility Question Need Help QUICK?
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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