Acid/Bases question please help

[TexasDDS]

buffer solutions....
"Add 0.1 mol of acetic acid (CH3COOH) and 0.1 mol of sodium acetate (CH3COONa) to water to obtain a 1- liter solution. Since acetic acid is a weak acid (Ka=1.75 x 10^-5), it will partially dissociate to give some acetate (CH3COO-) ions. However, the salt is soluble and will dissociate completely to give plenty of acetate ions. The addition of this common ion will shift the acid dissociation to the left, so the equilibrium concentrations of undissociated acetic acid molecules and acetate ions will be essentially equal to their initial concentrations, 0.1 M."
CH3COOH + H2O --> H3O+ + CH3COO-
It gives you the solution Ka = [H3O+][CH3COO-]/[ CH3COOH]

Which we can solve for [H3O+]
[H3O+]= Ka [CH3COOH]/[ CH3COO-]
Ka 0.1M/0.1 M = 1.75 x 10-5 --> pH = -log [H3O+] = -log(1.75 x 10-5) = 4.76
How did they get 4.76? I don't know how to get that without using a calculator can someone please explain… Thanks

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[AndyK]

buffer solutions....
"Add 0.1 mol of acetic acid (CH3COOH) and 0.1 mol of sodium acetate (CH3COONa) to water to obtain a 1- liter solution. Since acetic acid is a weak acid (Ka=1.75 x 10^-5), it will partially dissociate to give some acetate (CH3COO-) ions. However, the salt is soluble and will dissociate completely to give plenty of acetate ions. The addition of this common ion will shift the acid dissociation to the left, so the equilibrium concentrations of undissociated acetic acid molecules and acetate ions will be essentially equal to their initial concentrations, 0.1 M.”
CH3COOH + H2O --> H3O+ + CH3COO-
It gives you the solution Ka = [H3O+][CH3COO-]/[ CH3COOH]

Which we can solve for [H3O+]
[H3O+]= Ka [CH3COOH]/[ CH3COO-]
Ka 0.1M/0.1 M = 1.75 x 10-5 --> pH = -log [H3O+] = -log(1.75 x 10-5) = 4.76
How did they get 4.76? I don’t know how to get that without using a calculator can someone please explain… Thanks

-log(1e-5) = 5
-log(1e-4) = 4
so -log(1.75e-5) is somewhere between pH=4 and pH=5... it's closer to 5 because 1.75e-5 is closer to 1e-5 than to 1e-4 (which is like 10e-5) Using these rules of thumb, you will guess the right answer.

I don't think the DAT has these kinds of questions on it anyways.