1) a 50 ML solution of NaOH was titrated with 39.7 mL of .9781 M HCL. what is th
ID: 480963 • Letter: 1
Question
1) a 50 ML solution of NaOH was titrated with 39.7 mL of .9781 M HCL. what is the concentration of the NaOH solution
2)Twenty-five ML of .100 M lactic acid (ka=1.4×10^-4) is titrated with .097 M KOH.
a)What is the pH of the solution before titation?
b)How many milleters of KOH solution are required to titrate the acid?
c)what is the pH of the solution at the equivalence point?
3)if 50.00 ML of 1.00 M HC2H3O2 (Ka=1.8×10^-5) is titrated with 1 M NaOH. what is the pH of the solution after the following volumes of NaOH have been added?
a)0.00mL
b)25.00mL
c)49.90mL
d)50.10mL
Explanation / Answer
1)
mmoles of NaOH = 50 x M
mmoles of HCl = 39.7 x 0.9781 = 38.83
NaOH + HCl --------------> NaCl + H2O
from the balanced equation
moles of NaOH = moles of HCl
50 x M = 38.83
M = 0.7766
concentration of NaOH = 0.7766 M
2)
a)
HA --------------> H+ + A-
0.1 0 0
0.1-x x x
Ka = x^2 / 0.1 - x = 1.4×10^-4
x = 3.67 x 10^-3
[H+] = 3.67 x 10^-3 M
pH = -log [H+ ] = -log (3.67 x 10^-3 )
pH = 2.44
b)
mmoles of lactic acid = 25 x 0.1 = 2.5
mmoles of acid = mmoles of base
2.5 = 0.097 x V
volume of KOH = 25.77 mL
c)
pKa = 3.85
millimoles of salt = 2.5
concentration of salt = 2.5 / (25 + 25.77)
= 0.049 M
pH = 7 + 1/2 (pKa + log C)
= 7 + 1/2 (3.85 + log 0.049)
= 8.27
pH = 8.27
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