# 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?

1
2021-02-24T03:47:42-0500

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

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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