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Question: What is the pH of a 10.0 mL, 0.1 M \(N{{H}_{3}}\) solution when a 0.1 mL, 0.1 M NaOH solution was ad...

What is the pH of a 10.0 mL, 0.1 M NH3N{{H}_{3}} solution when a 0.1 mL, 0.1 M NaOH solution was added into it?

Explanation

Solution

pH is defined as the hydrogen potential of any solution. It has a range of 0 – 14. It tells us the acidic and basic nature of any solution. For acids it has a low value, and for base it has a higher value. The pH is calculated as the negative logarithm of the hydrogen ion concentration. From pOH value, the pH can be determined as, pH = 14 – pOH. pOH is the value of hydroxide ion potential.

Complete answer:
We have been given that 10.0 mL, 0.1 M NH3N{{H}_{3}}solution is mixed with a 0.1 mL, 0.1 M NaOH solution. So, the new concentration of NaOH in the solution will be calculated by the dilution formula as C1V1=C2V2{{C}_{1}}{{V}_{1}}={{C}_{2}}{{V}_{2}}, where C1{{C}_{1}}= 0.1 mol/L, V1{{V}_{1}}=0.1mL, and V2{{V}_{2}}= (10 + 0.1)mL, so,
C2=C1×V1V2{{C}_{2}}={{C}_{1}}\times \dfrac{{{V}_{1}}}{{{V}_{2}}}
C2=0.1 mole/L×0.10mL10.1mL{{C}_{2}}=0.1\,mole/L\times \dfrac{0.10mL}{10.1mL}
C2=9.9×10−4mol/L{{C}_{2}}=9.9\times {{10}^{-4}}mol/L
Now, according to the equation, NH3+H2O→NH4++OH−N{{H}_{3}}+{{H}_{2}}O\to N{{H}_{4}}^{+}+O{{H}^{-}}
Where initial and final concentrations are,
                          NH3+H2O→NH4++OH− initial (M)      0.1                           0               9.9×10−4 final (M)        0.1−x                   x             x+9.9×10−4 \begin{aligned} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,N{{H}_{3}}+{{H}_{2}}O\to N{{H}_{4}}^{+}+O{{H}^{-}} \\\ & initial\,(M)\,\,\,\,\,\,0.1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,9.9\times {{10}^{-4}} \\\ & final\,(M)\,\,\,\,\,\,\,\,0.1-x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,x+9.9\times {{10}^{-4}} \\\ \end{aligned}
Now, according to the dissociation constant of a base, this is the value of the concentration of the products upon the reactant, Kb=[NH4+][OH−][NH3]=1.8×10−5{{K}_{b}}=\dfrac{[N{{H}_{4}}^{+}][O{{H}^{-}}]}{[N{{H}_{3}}]}=1.8\times {{10}^{-5}} . From this value the value of x can be calculated as
Kb=x(x+9.9×10−4)0.1−x=1.8×10−5{{K}_{b}}=\dfrac{x(x+9.9\times {{10}^{-4}})}{0.1-x}=1.8\times {{10}^{-5}}, solving this for x, we have the value of x as
x=9.9×10−4x=9.9\times {{10}^{-4}}
Now, we have the hydroxide ion concentration as [OH−]=9.9×10−4+x[O{{H}^{-}}]=9.9\times {{10}^{-4}}+x
[OH−]=9.9×10−4+9.9×10−4[O{{H}^{-}}]=9.9\times {{10}^{-4}}+9.9\times {{10}^{-4}}
[OH−]=2.0×10−3M[O{{H}^{-}}]=2.0\times {{10}^{-3}}M
Keeping this in the formula for, pOH = −log⁡[OH−]-\log [O{{H}^{-}}], we have
pOH=−log⁡(2.0×10−3)pOH=-\log (2.0\times {{10}^{-3}})
pOH = 2.7
Now using, pH = 14 – pOH we have,
pH = 14 – 2.7
pH = 11.3
hence, the pH of the solution is found to be 11.3 which is a basic pH.

Note:
pOH is the negative log of hydroxide ion concentration, it is calculated for bases. The dissociation constant for acid is Ka{{K}_{a}} while for base is Kb{{K}_{b}}. The value of Ka×Kb=10−4{{K}_{a}}\times {{K}_{b}}={{10}^{-4}}. The ammonia gas and NaOH does not react usually, here we are dissolving the ammonia into water to form ammonium ion and the hydroxide ions that takes the place of the base NaOH, as we add NaOH the hydroxide ions increase in this solution.