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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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nevada nevada
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11 years ago
A solution is made by adding 50.0 mL of 0.200 M acetic acid (Ka =1.8 x 10^-5) to 50.0 mL of 1.00 x 10^-3 M HCl.
a. Calculate the pH of the solution.
b. Calculate the acetate ion concentration.

I don't know how to start this problem, I know it's has to be a ice chart  because if the weak acid.
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#1
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11 years ago
HCl is strong acid
it will dissociate completely, thus concntration contributed by HCl
= 0.001*50/100 =0.0005 M
CH3COOH - > H+ CH3COO(-)
new concentration after mixing, Acetic acid = 50*0.2/100 =0.1 M
CH3COOH - > H+ CH3COO(-)
0.1
0.1-X X+0.0005 X
Now,

Ka = (X+0.0005)*(X)/(0.1-X) = 1.8*10^(-5)
X=0.0011 M
Hence total [H+] = 0.0005+0.0011 =0.0016 M
Hence, pH = -log(H) = -log(0.0016) =2.795 ~ 2.8
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