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    Ionization

    Updated7 min read

    4. 49 gm of H2SO4 is present in 1000 ml of its solution. What is the pH of the solution?
    solution:

    Molarity=\frac{gm/lit}{mol.\ wt.}=\frac{49}{98}=0.5M\\ \underset{\underset{0.5M}{1M}}{H_{2}SO_{4}}\rightarrow \underset{\underset{1M}{2M}}{2H^{+}}+SO_{4}^{--}\\ \therefore pH=-log[H^{+}]=-log[1]=0

    5. Calculate the pH of 0.1N H2SO4?
    solution:

    Molarity=\frac{Normality}{Basicity}=\frac{0.1}{2}=0.05M\\ \\ \underset{\underset{0.5M}{1M}}{H_{2}SO_{4}}\rightarrow \underset{\underset{0.1M}{2M}}{2H^{+}}+SO_{4}^{--}\\ \therefore pH=-log[H^{+}]=-log[0.1]=1

    6. 10 cc of N/2 HCl, 30 cc of N/10 HNO3 and 60 cc of N/5 H2SO4 are mixed together, calculate the pH of the mixture.
    solution:

    HCl solution
    S1 = N/2
    V1 = 10 cc    		HNO3 solution
    S2 = N/10
    V2 = 30 cc    		H2SO4 solution
    S3 = N/5
    V3 = 60 cc
    \begin{align*}so,\ V_{mix}S_{mix}&=V_{1}S_{1}+V_{2}S_{2}+V_{3}S_{3}\\ S_{mix}&=\frac{V_{1}S_{1}+V_{2}S_{2}+V_{3}S_{3}}{V_{mix}}\\ &=\frac{10\times 1/2\ +\ 30\times 1/10\ +\ 60\times1/5}{10+30+60}\\ &=0.2\\ Now,\ [H^{+}]&=0.2\ mol/lit\\ pH&=-log[H^{+}]=-log[0.2]=0.696 \end{align*}

    7. Calculate the pH of a mixture obtained by mixing 75 ml of 0.2M HCl and 25 ml of 0.2 M NaOH solution.
    solution:

    HCl solution
    V1 = 75ml
    S1 = 0.2M = 0.2N    		NaOH solution
    V2 = 25 ml
    S2 = 0.2M = 0.2 N
    \begin{align*}so,\ V_{mix}S_{mix}&=V_{1}S_{1}-V_{2}S_{2}\\ S_{mix}&=\frac{V_{1}S_{1}-V_{2}S_{2}}{V_{mix}}\\ &=\frac{75\times 0.2\ +\ 25\times 0.2}{75+25}\\ &=0.1\\ Since,\ & HCl\ is\ in\ excess, [H^{+}]=0.1\\ pH&=-log[0.1]=1 \end{align*}

    8. 200 ml of an aqueous solution of HCl (pH=2) is mixed with 300 ml of an aqueous solution of NaOH (pH=12). What will be the pH of the resulting mixture solution?
    solution:

    For HCl,
    Volume = 200 ml
    pH = -log[H+] or [H+] = antilog[-pH]
    = antilog[-2] = 0.01M = 0.01N    		For NaOH,
    Volume = 300 ml
    pH + pOH = 14 or, pOH = 2
    or, [OH–] = antilog[-2] = 0.01M = 0.01N

    For mixture,

    \begin{align*}Volume&=200+300=500ml\\ V_{2}N_{2}-V_{1}N_{1}&=V_{m}N_{m}\\ N_{m}&=\frac{300 \times 0.1\ - 200 \times 0.1}{500}\\ &= 2 \times 10^{-3}M\\ \underset{2 \times 10^{-3}M}{NaOH} &\rightarrow Na^{+} + \underset{2 \times 10^{-3}}{OH^{-}}\\ [OH^{-}]&=2 \times 10^{-3}M\\ \because [H^{+}][OH^{-}]&=10^{-14}\\ or,\ [H^{+}]&=5 \times 10^{-12}M\\ or,\ pH&=-log[H^{+}]=-log[5 \times 10^{-12}]\\ \therefore pH&=11.3 \end{align*}

    9. Find the pH of 0.01M acetic acid solution. Ka = 1.7 x 10-5

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