what is the ph of the solution. obtained by mixing 50 ml of .1250 M hoax 4. What
ID: 972347 • Letter: W
Question
what is the ph of the solution. obtained by mixing 50 ml of .1250 M hoax 4. What is the pH of the solution obtalned by mixing 50.00 mL of 0.1250 M HOAc and 25.00 mL of 0.1000 M NaOH? 5) Calculate the pH of a solution prepared by mixing 10.0 500 M Na0H and 20.0 mL of 0.50 M ben of zolc acld solution. (Benzolc ' acid is monoprotle: Its lonization constant is 6.7 x 10-5.) If Ka ls 1.85x 10-s for acetic acld, calculate the pH at one-half the equivalence polnt and at the equivalence point for a titration of 50 ml of o.10oM acet ic acid with 0.100 H NaoH. (HInt: hydrolysis is possible)Explanation / Answer
pKa of acetic acid = 4.756
mol of CH3COOH = M*V= 0.1250 M*0.05 L = 6.25*10^-3 mol
mol of NaOH = M*V= 0.1000 M*0.025 L = 2.5*10^-3 mol
2.5 *10^-3 mol of each will react to form 2.5*10^-3 mol of CH3COONa
finally,
total volume = 0.05+0.025 = 0.075 L
[CH3COONa] = 2.5*10^-3 mol / 0.075 L = 0.033 M
[CH3COOH] = (6.25*10^-3 - 2.5*10^-3)/0.075 = 0.05 M
use:
pH = pKa + log {[CH3COONa]/[CH3COOH]}
= 4.756 + log {0.033/0.05}
= 4.576
Answer: 4.576
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