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1)I found the answer for the first half of the problem, but need help with part two. A buffer solution that is 0.10 M sodium acetate and 0.20 M acetic acid is prepared. Calculate the initial pH of this solution. The Ka for CH3COOH is 1.8 x 10-5 M. As usual, report pH to 2 decimal places. Answer: 4.44 Correct A buffered solution resists a change in pH. Calculate the pH when 21.5 mL of 0.044 M HCl is added to 100.0 mL of the above buffer.
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Well, use M\(_1\)V\(_1\) = M\(_2\)V\(_2\) to fnd the second molar concentration, and then use it to find pH. pH = -log[H\(^+\)]
thank you!
so would the answer be 2.024? I set up the equation like this. .0215L*.044M = .1L * M2 THEN I GOT 9.46e-3 and took the -log of that.
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