Question

An aqueous solution contains 0.01 M RNH 2
​ (K b
​ =2×10 −6
) and 10 −4
M NaOH.The concentration of OH −
is nearly:

Answer 1

Question:

An aqueous solution contains 0.01 M $\rm RNH_2$ ($\rm K_b=2 \times 10^-6$) and $\rm 10^-4$ M NaOH.

The concentration of $\rm OH^-$ is nearly.

Answer:

Given:

⇒ 0.0.1 M $\rm RNH_2$

⇒ $\rm K_b=2 \times 10^-6$

⇒ $\rm 10^-4\;M\;NaOH$

To Find:

The concentration of $\rm OH^-$

Solution:

We know that,

$\rm \rm RNH_2$ is a weak base.

Weak base means on dissolution it does disassociate completely.

$\rm RNH_2$ reacts with Water ($\rm H_2O$) to give $\rm OH^-$ ions

The Formula to be used here is

$\green{\boxed{\rm [OH^-]=\sqrt{K_c \times C}+[NaOH] }}$

Given that

$\implies \rm K_b=2 \times 10^-6$

$\implies \rm C=0.01\;M=10^-2$

$\implies \rm [NaOH]=10^-4$

Substituting the above values in the formula

We get,

$\rm [OH^-]=\sqrt{2 \times 10^-6 \times 10^-2}+10^-4$

$\rm \implies 1.41 \times 10^-4 + 10^-4$

$\rm \implies(1.41+1) \times 10^-4$

$\rm \implies2.41 \times 10^-4$

Hence,

The concentration of $\rm OH^-$ is $\rm 2.41 \times 10^-4\;M$