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

Byju's Answer

Standard XII

Chemistry

Solubility Product

A sample of h...

Question

A sample of hard water contains $0.005$ mole of $C a {C l}_{2}$ per litre. What is the minimum concentration of ${N a}_{2} {S O}_{4}$ which must be exceeded for removing calcium ions from their water solution? (The solubility product of $C a {S O}_{4}$ is $2.4 \times 10^{- 5}$ .)

Open in App

Solution

The solubility product of $C a S O_{4}, K_{s p} = 2.4 \times 10^{- 5}$
$∴ [C a^{2 +}] = 0.005 M$
$\Rightarrow K_{s p} = [C a^{2 +}] [S O_{4}^{2 -}]$
$\Rightarrow [S O_{4}^{2 -}] = \frac{2.4 \times 10^{- 5}}{0.005} = 4.8 \times 10^{- 3} M$
$∴$ The minimum concentration of $N a_{2} S O_{4}$ will be $4.8 \times 10^{- 3} M$ .

Suggest Corrections

Similar questions

Q. A sample of hard water contains 0.005 moles of

C a C l_{2}

per litre. What is the minimum concentration Of

N a_{2} S O_{4}

which must be added for removing

C a^{2 +}

ions from the water sample?

[

K_{s p}

for

C a S O_{4}

2.4 \times 10^{- 5}

25^{\circ} C

]

Q. A sample of hard water contains

0.005 m o l e

of calcium chloride

p e r l i t r e

. What is the minimum concentration of sodium sulphate which must be added for removing the

{C a}^{2 +}

ions from this water sample?

(K_{s p}

C a {S O}_{4} = 2.4 \times 10^{- 5})

Q. The solubility product of

C a {S O}_{4}

2.4 \times 10^{- 5}

. When

100 m L

0.01 M

C a {C l}_{2}

and

100 m L

0.02 M

{N a}_{2} {S O}_{4}

are mixed then:

Q. One litre of a sample of hard water contains 11.1 g

C a C l_{2}

9.5 g M g C l_{2}

. Find the concentration of

C l^{-}

ion in the solution.

You are provided with a $500 m L$ of hard water, containing $0.005 m o l e$ of $C a C l_{2}$ , and two sulphuric acid samples of $0.001 M$ and $0.02 M$ concentrations. The solubility product of calcium sulphate in water at $25^{\circ} C$ is $2.4 \times 10^{- 5}$ . Choose the most appropriate answer:

Join BYJU'S Learning Program

A sample of hard water contains 0.005 mole of CaCl2 per litre. What is the minimum concentration of Na2SO4 which must be exceeded for removing calcium ions from their water solution? (The solubility product of CaSO4 is 2.4×10−5.)

The solubility product of CaSO4,Ksp=2.4×10−5∴[Ca2+]=0.005M⇒Ksp=[Ca2+][SO2−4]⇒[SO2−4]=2.4×10−50.005=4.8×10−3M∴ The minimum concentration of Na2SO4 will be 4.8×10−3M .

A sample of hard water contains $0.005$ mole of $C a {C l}_{2}$ per litre. What is the minimum concentration of ${N a}_{2} {S O}_{4}$ which must be exceeded for removing calcium ions from their water solution? (The solubility product of $C a {S O}_{4}$ is $2.4 \times 10^{- 5}$ .)