You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Stoichiometric Calculations

The phosphoru...

Question

The phosphorus in a $4.258$ g sample of a plant food was converted to ${P O_{4}}^{3 -}$ and precipitated as $A g_{3} P O_{4}$ through the addition of $50.0$ mL of $0.082 M A g N O_{3}$ . The excess of $A g N O_{3}$ was back-titrated with $4.86$ mL of $0.0625 M$ $K S C N$ . Express the results of this analysis in terms of $P_{2} O_{5}$ as nearest integer $%$ .
$P_{2} O_{5} + 9 H_{2} O \to 2 {P O_{4}}^{3 -} + 6 {H_{3} O}^{+}$
$2 {P O_{4}}^{3 -} + 6 A g^{+} \to 2 A g_{3} P O_{4}$
$A g^{+} + S C N^{-} \to A g S C N (s)$

Open in App

Solution

Let $V$ mL represent the volume of unreacted $A g N O_{3}$ .
$V$ mL of $0.082 M$ $A g N O_{3}$ corresponds to $4.86$ ml of $0.0625 M$ $K S C N$ .
$V = \frac{4.86 \times 0.0625}{0.082} = 3.70$ mL
The number of moles of $A g N O_{3}$ used $\frac{46.30 \times 0.082}{1000} = 3.8 \times 10^{- 3}$

$P_{2} O_{5} \to 2 {P O_{4}}^{3 -}$
$2 {P O_{4}}^{3 -} + 6 A g^{+} \to 2 A g_{3} P O_{4}$

The $P_{2} O_{5}$ content is $\frac{3.8 \times 10^{- 3}}{6} = 6.33 \times 10^{- 4} = \frac{3.8 \times 10^{- 3}}{6} \times 142 = 0.0898 g$

The percentage of $P_{2} O_{5}$ present in plant food is $\frac{0.0898}{4.258} \times 100 = 2.11$ $\sim 2 %$

Suggest Corrections

Similar questions

Q. A

0.141

g sample of phosphorus containing compound was digested in a mixture of

H N O_{3}

and

H_{2} S O_{4}

, which resulted in formation of

C O_{2}, H_{2} O

and

H_{3} P O_{4}

. Addition of ammonium molybdate yielded a solid having the composition

(N H_{4})_{3} P O_{4} .12 M o O_{3}

. The precipitate was filtered, washed and dissolved in

50.0

mL of

0.20 M N a O H

(N H_{4})_{3} P O_{4} .12 M o O_{3} (s) + 26 O H^{-} ⟶ H P O_{4}^{2 -} + 12 M o O_{4}^{2 -} + 14 H_{2} O

+ 3 N H_{3} (g)

After boiling the solution to remove the

N H_{3}

, the excess of

N a O H

was titrated with

14.1

mL of

0.174 M

H C l

. Calculate the percentage of phosphorus in the sample.

Q. A sample of

P_{2} O_{6}

contains some

H_{3} P O_{4}

impurity. A

0.405

g sample is reacted with water (

P_{2} O_{5} + 3 H_{2} O \to 2 H_{3} P O_{4}

), and the resulting solution is titrated with

0.25 M

N a O H

requiring

42.5

mL (upto

N a_{2} H P O_{4}

). The percentage purity of

H_{3} P O_{4}

as nearest integer is :

Q. Titration of a

0.7439

g sample of impure

N a_{2} B_{4} O_{7} .10 H_{2} O

(borax) requires

31.64

mL of

0.108 M H C l

for the reaction. The result of this analysis in terms of percentage is (in nearest integer) :

(Molecular weight of

B = 11

g/mol)

Q. A

1.5

g sample containing oxalic acid and some inert impurity was dissolved in enough water and volume was made up to

250

mL. A

20

mL portion of this solution was then mixed with

30

mL of an alkali solution. The resulting solution was then treated with stoichiometric amount of

C a C l_{2}

just needed for precipitation of oxalate as

C a C_{2} O_{4}

solution was filtered off and filtrate was finally titrated against

0.1 M H C l

solution.

8.0

mL of acid was required to reach the equivalence point. At last the above neutral solution was treated with excess of

A g N O_{3}

solution and

A g C l

obtained was washed, dried and weighed to be

0.4305

g. Determine the mass percentage of oxalic acid in the original sample.

5

g of

K_{2} S O_{4}

was dissolved in

250

mL of solution. The amount of mL of this solution that should be used so that

1.2

B a S O_{4}

may be precipitated by this solution on addition of

B a {C l}_{2}

solution is __________. (as the nearest integer)

Join BYJU'S Learning Program