It's difficult to interpret with improper spacing. Perhaps this is better. I'll do the first one and leave the others for you. Post your work if you get stuck. I'll call the strong acid HA and the strong base BOH.
.........Acid.........................................................................Base
Volume (L) .............Conc. [M] ..................Volume (L)............ Conc. [M]
a)... 1.760.................. 0.384....................... 0.834...................... 0.724
b) 0.493 1.20 0.375 1.20
c) 0.038 0.0055 0.293 0.0010
d) 0.294 0.035 0.015 0.38
e) 5.013 5.7×10-4 0.395 3.4×10-1
.....................HA + BOH ==> AB + H2O
mols HA = M x L = 1.760 x 0.384 = 0.676
mols BOH = M x L = 0.834 x 0.724 = 0.604
mols HA left in excess is 0.676 - 0.604 = 0.0720 mols in a volume of 1.760 L + 0.834 L = 2.59 L so concentration of the HA remaining is M = mols/L = 0.0720 mols/2.59 L = 0.0278 M. pH = -log(H^+) = -log(0.0278) = 1.56
Use the following table of volumes and concentrations of strong acids and bases to determine the final pH for each solution (a - e). Use three significant figures in your answer.
Acid Base
Volume (L) Conc. [M] Volume (L) Conc. [M]
a) 1.760 0.384 0.834 0.724
b) 0.493 1.20 0.375 1.20
c) 0.038 0.0055 0.293 0.0010
d) 0.294 0.035 0.015 0.38
e) 5.013 5.7×10-4 0.395 3.4×10-1
2 answers
Can I assume that the acid is monoprotic?
2 answers