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 H_{2}SO_{4},

molar equivalent mass=49.04g/eq

Molar mass = 98.08g/mol

Thus;

Molarity of H_{2}SO_{4} "=\\dfrac{1.0923Nx49.04g\/eq}{98.08g\/mol}=0.54615M"

Moles of H_{2}SO_{4} "=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) + H_{2}SO_{4}(aq) "\\to" ZnSO4(aq) + H2O(l)

Moles of H_{2}SO_{4} that reacted with ZnO;

Mole ratio of ZnO:H2SO4 = 1:1

Moles of H_{2}SO_{4} "=\\dfrac{1}{1}x0.0182mol = 0.0182mol"

Moles of H_{2}SO_{4} that remained in the solution;

=0.0273mol - 0.0182mol = 0.0091mol

III. Equation for the back titration

H_{2}SO_{4(aq)} + 2NaOH_{(aq)} "\\to" Na_{2}SO_{4(aq)} + 2H_{2}O_{(l)}

Mole ratio of H_{2}SO_{4} : 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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