The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Advertisement Remove all ads
Solution
Solubility of Sr(OH)2 = 19.23 g/L
Then, concentration of Sr(OH)2
`= 19.23/121.63 M`
= 0.1581 M
`Sr(OH)_(2(aq)) -> Sr_((aq))^(2+) + 2(OH^(-))_((aq))`
`:. [Sr^(2+)] = 0.1581 M`
`[OH^(-)] = 2xx 0.1581 M = 0.3126 M`
Now
`K_w = [OH^-][H^+]`
`10^(-14)/0.3126 = [H^+]`
`=> [H^+] = 3.2 xx 10^(-14)`
`:. pH = 13.495; 13.50`
Is there an error in this question or solution?
Advertisement Remove all ads
APPEARS IN
Advertisement Remove all ads
Advertisement Remove all ads
