Answer to Question #165664 in General Chemistry for Kei

Question

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?

Answer 1

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

Thus;

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

H₂SO_4(aq) + 2NaOH_(aq) $\to$ Na₂SO_4(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?

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