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

Solubility Product

The solubilit...

Question

The solubility of ${P b S O}_{4}$ water is $0.0608 g / L$ . Calculate the solubility product constant of ${P b S O}_{4}$ .
Molar mass ${P b S O}_{4} = 304 g / m o l e$

Open in App

Solution

The solubility of $P b S O_{4}$ is $0.0608 g / L$
Number of moles $=$ $\frac{m o l e s}{R e l a t i v e f o r m u l a m a s s}$
$\Rightarrow \frac{0.0608}{304} = 0.0002$ moles
$K_{s p} = [P b^{2 +}] [S O_{4}^{2 -}]$
$\Rightarrow (0.0002)^{2} = 4 \times 10^{- 8}$ .

Suggest Corrections

Similar questions

P b S O_{4}

P b S O_{4}

P b S O_{4} = 304 g / m o l e \frac{380}{304} = 1.25 (1.25)^{2} = 1.56

P b {S O}_{4}

P b {S O}_{4}

{P b S O}_{4}

0.03 g / L

{P b S O}_{4}

(K = {\times 10}^{- 12}

1.0 L

0.010 M N a l

P b {S O}_{4}

0.038 g

L^{- 1}

25^{o} C

P b {S O}_{4}

0.01 M

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

K_{s p}

P b {S O}_{4} = 1.25 \times 10^{- 9}

Join BYJU'S Learning Program