If pH of a saturated solution of textBa( textOH)2 is 12, the value of its K(. textsp .) is found to be n× 10- 7M3. find the value of ‘n’ here?

Question

If pH of a saturated solution of textBa( textOH)2 is 12, the value of its K(. textsp .) is found to be n× 10- 7M3. find the value of ‘n’ here? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

As textpH= text12 textpOH= text14- text12=2 [. textOH].=10- 2M=0.0 text1M/L In case of Ba(OH)2 leftharpoons Ba+2 + 2OH-2 As text2s=0.0 text1m / L s=(0.01/2)=5× 10- 3m / L mathrmKsp for mathrmBa( mathrmOH)2=4 mathrm~s3=4(5 × 10-3)3=5 × 10-7 mathrmM3 Hence 'n' = 5

Q. If pH of a saturated solution of Ba(OH)2​ is 12, the value of its K(sp)​ is found to be n×10−7M3. find the value of ‘n’ here?

As pH=12 pOH=14−12=2 [OH]=10−2M=0.01M/L In case of Ba(OH)2​⇌Ba+2+2OH−2 As 2s=0.01m/L s=20.01​=5×10−3m/L Ksp for Ba(OH)2​=4 s3=4(5×10−3)3=5×10−7M3 Hence ′n′ = 5

Q. If pH of a saturated solution of $Ba (OH)_{2}$ is $12,$ the value of its $K_{(sp)}$ is found to be $n \times 1 0^{- 7} M^{3} .$ find the value of $‘ n ’$ here?