Question

the ph of solution 0.1M NH3. Kb = 1.8 x 10^-5

Answer 1

we have to find ka as kb is given it can be found by using pKA plus p KB is equal to pKW as we know it is 14 from there you'll get pKa as 9.25 from that a is equal to 5.6 into 10 power minus 10 from there you can find the h + Ion concentration and substitute in pH you get the pH of H Plus

Answer 2

Answer:

The pH of the solution of $0.1 M$ $NH_{3}$ measured is $11.12$.

Explanation:

Given,

The dissociation constant, $K_{b}$ = $1.8\times 10^{-5}$

The concentration of $NH_{3}$ = $0.1M$

The pH of the solution =?

As we know,

• The reaction of $NH_{3}$ with the aqueous solution follows in the way given below:
• $NH_{3} + H_{2}O \rightleftarrows NH_{4}^{+} +OH^{-}$

• $0.1$                     x         x

Now,

• The equilibrium constant, $K_{b}$ = $\frac{Concentration of product}{Concentration of reactant}$ = $\frac{[NH_{4}^{+}] [OH^{-}] }{[NH_{3}] }$
• $1.8\times 10^{-5}$ = $\frac{x^{2} }{0.1}$
• $x^{2} = 1.8\times 10^{-6}$
• x = $1.34\times 10^{-3}$

Now, the concentration of $[OH^{-}]$ = $1.34\times 10^{-3}$

As we know,

• pOH = $-log[OH^{-}]$
• pOH = $-log[1.34\times 10^{-3}]$
• pOH = $-[(log 1.34) + (-3log10)]$
• pOH = $-[0.127 -3]$
• pOH = $2.87$

Also,

• $pOH + pH = 14$
• $2.87 + pH = 14$
• pH = $11.12$

Hence, the pH of the solution = $11.12$.