Answer to Question #165664 in General Chemistry for Kei

Question #165664

If a 1.54g sample of zinc oxide,

96.2%, were treated with 50ml of

1.0923N H2SO4 in the usual way,

what volume of 0.9765N sodium

Hydroxide would be required in the

back titration?

Expert's answer

I. First we calculate the actual mass of ZnO

Actual mass "=" "\\dfrac{96.2}{100}x1.54g = 1.48148g"

Then the moles of ZnO will be;

moles "=" "\\dfrac{mass}{MM} = \\dfrac{1.48148g}{81.38g\/mol}=0.0182mol"

II. We then need to convert Normality into Molarity for easy calculations of moles of the solutions

Since Normality "=\\dfrac{MolarityxMolar Mass}{Equivalent mass}"

Molarity "=\\dfrac{Normalityxequivalent mass}{Molar Mass}"

For 1.0923N H2SO4,

molar equivalent mass=49.04g/eq

Molar mass = 98.08g/mol


Molarity of H2SO4 "=\\dfrac{1.0923Nx49.04g\/eq}{98.08g\/mol}=0.54615M"

Moles of H2SO4 "=Molarity x Volume=0.54615mol x \\dfrac{50mL}{1000mL}= 0.0273 moles"

For NaOH;

Molarity "=\\dfrac{0.9765N}{1eq} = 0.9765M"

Equation for the reaction between ZnO and H2SO4;

ZnO(s) + H2SO4(aq) "\\to" ZnSO4(aq) + H2O(l)

Moles of H2SO4 that reacted with ZnO;

Mole ratio of ZnO:H2SO4 = 1:1

Moles of H2SO4 "=\\dfrac{1}{1}x0.0182mol = 0.0182mol"

Moles of H2SO4 that remained in the solution;

=0.0273mol - 0.0182mol = 0.0091mol

III. Equation for the back titration

H2SO4(aq) + 2NaOH(aq) "\\to" Na2SO4(aq) + 2H2O(l)

Mole ratio of H2SO4 : NaOH = 1:2

Moles of NaOH that would react;

"=\\dfrac{2}{1}x0.0091mol = 0.0182mol"

IV. Now, calculate volume of NaOH from moles

0.9765 moles were in 1000mL of NaOH

0.0182 moles will be in what volume?

"=\\dfrac{0.0182mol x 1000mL}{0.9765mol}= 18.64mL of NaOH"

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