Calculate the pH’s of the following: (a) the solution formed by mixing 20.0 mL o
ID: 507499 • Letter: C
Question
Calculate the pH’s of the following: (a) the solution formed by mixing 20.0 mL of a 0.30M HCl solution with 12 mL of a 0.25 M NaOH. (b) the solution formed by mixing 18.0 mL of a 0.46M HC6H7O6 solution with 10.0 mL of a 0.35M NaOH (Ka for HC6H7O6 = 8.0x10 –5 ) (c) the solution formed by mixing 15.0 mL of a 0.47M HF with 20.0 mL of a 0.20M KOH solution (Ka for HF = 7.1x10 –4 ) (d) the solution formed by mixing 30.0 mL of a 0.20 M HCHO2 solution with 22.0 mL of a 0.40 M KOH solution (Ka for HCHO2 = 1.7x10 –4 ) (e) the solution formed by dissolving 2.85 g of KF in 120 mL of a 0.30 M HF solution (Ka for HF = 7.1x10 –4 ) (f) the end point of the titration of 58.0 mL of a 0.48 M HCHO2 with 0.40 M KOH.
Explanation / Answer
HCl NaOH
MA = 0.3M MB = 0.25M
VA = 20ml VB = 12ml
M = MAVA-MBVB/VA+ VB
= 0.3*20-0.25*12/20+12 = 3/32 = 0.09375M
M = [H+] = 0.09375M
PH = -log[H+]
= -log0.09375 = 1.028
b.
no of moles of HC6H7O6 = molarity * volume in L
= 0.46*0.018 = 0.00828moles
no of moles of NaOH = molarity * volume in L
= 0.35*0.01 = 0.0035moles
Pka = -logKa
= -log8*10^-5
= 4.0969
HC6H7O6 + NaOH -----------> NaC6H7O6 + H2O
I 0.00828 0.0035 0
C -0.0035 -0.0035 0.0035
E 0.00478 0 0.0035
PH = Pka + log[ NaC6H7O6]/[ HC6H7O6]
= 4.0696 + log0.0035/0.00478
= 4.0696-0.1353 = 3.9353
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