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All Message Boards : Acid and Base
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 Message 1 of 2 in Discussion 
From: MSN NicknameAlbert1145  (Original Message)Sent: 9/22/2008 7:53 AM
Hi Steve, I've been doing a lot of acid and base problems lately. They seem pretty easy but I have two question that I don't get:

A buffer solution is prepared by taking 0.400 moles of acetic acid (pKa = 4.74) and
0.250 moles of sodium acetate in sufficient water to make 1.400 L of solution. What is
the pH of this solution?
(A) 4.46
(B) 4.56
(C) 4.86
(D) 4.96
ANS: (C) but I keep on getting pH 4.53

A mixture is made using 50.0 mL of 0.20 mol L�? HCl(aq) and 150.0 mL of distilled
water. What is the pH of this solution? I think I'm having some trouble with the unit conversion in this one... do I actually need the concentration of distilled water to calculate this question??
(A) 1.30
(B) 13.00
(C) 6.48
(D) 9.16
ANS: (A)

Thank you


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Reply
 Message 2 of 2 in Discussion 
From: MSN Nickname·Steve·Sent: 9/22/2008 7:36 PM
Hi Albert, your answer to the first problem is correct. I get pH = 4.74 + log(0.250 mol / 0.400 mol) = 4.74 �?0.204 = 5.54.
 
For the second problem, just plug into the dilution formula to get the molarity of the diluted solution of HCl:
 
VconcMconc  =  VdilMdil
(50.0 mL)(0.20 M)  =  (200 mL)Mdil 
Mdil = 0.050 M
 
Since HCl is a strong acid, the H+ concentration is the same as the molarity, 0.050 M. Thus the pH is just –log(0.050 M) = 1.30. If you had a very dilute solution of HCl, then the concentration of H+ ion in the water itself would be significant (1.0 X 10�? M at 25°C) but it would be unusual to have to consider this.
 
 
Steve