A volume of 70.0 mL of aqueous potassium hydroxide (KOH) was titrated against a

Question

A volume of 70.0 mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4)What was the molarity of the KOH solution if 24.2 mL of 1.50 M H2SO4 was needed? The equation is 2KOH(a) +H2SO4(aq) KoSO4(a)2H2Ol) Express your answer with the appropriate units. Calculate the percent ionization offormic acid (???2H) in a solution that is 0.152 M in formic acid. The Ka of formic acid is 1.77-10-4. ? 8.44 2.74-105 0.0180 3.44 The average human body contains 6.00 L of blood with a Fe2+ concentration of 2.60x10-5 M. If a person ingests 8.00 mL of 11.0 mMNaCN, what percentage of iron(lI) in the blood would be sequestered by the cyanide ion?

Explanation / Answer

1.

24.2 mL of 1.50 M H2SO4.

So, moles of H2SO4 = 1.50 M x 0.0242 L = 0.0363 moles

Now,

2KOH(aq)   +   H2SO4(aq)   ------>   K2SO4(aq)   +   2H2O(l)

In the above reaction;
     1 mole of H2SO4 reacts with 2 moles of KOH
or, 0.0363 mole of H2SO4 reacts with (2 x 0.0363) moles of KOH
or, 0.0363 mole of H2SO4 reacts with 0.0726 moles of KOH

Volume of KOH = 70 mL = 0.070 L

So, molarity of KOH = 0.0726 moles / 0.070 L = 1.0371 M

2.

      HCOOH       =        H+        +        HCOO-
IC:             0.152                      0                        0
C:                 - x                     + x                     + x
EC:          0.152 – x                   x                       x

So, Ka = [HCOO-] [H+] / [HCOOH]

1.77 x 10-4 = (x) (x) / (0.152 – x)

Or, 1.77 x 10-4 = x2 / (0.152 – x)

Or, 1.77 x 10-4 = x2 / 0.152    (Ka is very small, so, x term in denominator is neglected)

Or, x2 = (1.77 x 10-4) x 0.152

x2 = 2.69 x 10-5

x = 0.0052

So, percentage of ionization = (0.0052 /0.152) x 100
                                             = 3.42

Answer is option (d) 3.44

3.

moles of Fe2+ = (6.00 L) x (2.60 x 10^-5 M) = 0.000156 moles

moles of CN- = (0.008 L) x (0.0110 M) = 0.000088 moles

Fe2+ + 6 CN- = [Fe(CN)6]4-

moles Fe2+ sequestered = 0.000088 / 6 = 0.0000147

% = (0.0000147 / 0.000156) x 100 = 9.42

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