A buffer solution contains 0.475 M nitrous acid and 0.302 M sodium nitrite . If 0.0224 moles of potassium hydroxide are added to 150 mL of t

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A buffer solution contains 0.475 M nitrous acid and 0.302 M sodium nitrite . If 0.0224 moles of potassium hydroxide are added to 150 mL of this buffer, what is the pH of the resulting solution

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Mộc Miên 2 weeks 2021-07-22T07:06:54+00:00 1 Answers 2 views 0

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    2021-07-22T07:08:04+00:00

    Answer: The pH of the resulting solution will be 3.001

    Explanation:

    Molarity is calculated by using the equation:

    \text{Molarity}=\frac{\text{Moles}}{\text{Volume}}          ……(1)

    We are given:

    Moles of NaOH = 0.0224 moles

    Molarity of nitrous acid = 0.475 M

    Molarity of sodium nitrite = 0.302 M

    Volume of solution = 150 mL = 0.150 L          (Conversion factor: 1 L = 1000 mL)

    Putting values in equation 1, we get:

    \text{Moles of nitrous acid}=(0.475mol/L\times 0.150L)=0.07125mol

    \text{Moles of sodium nitrite}=(0.302mol/L\times 0.150L)=0.0453mol

    The chemical equation for the reaction of nitrous acid and NaOH follows:

                       HNO_2+NaOH\rightleftharpoons NaNO_2+H_2O

    I:                0.07125     0.0224     0.0453

    C:             -0.0224     -0.0224   +0.0224

    E:              0.04885         –          0.0677

    The power of the acid dissociation constant is the negative logarithm of the acid dissociation constant. The equation used is:

    pK_a=-\log K_a        ……(2)

    We know:

    K_a for nitrous acid = 7.2\times 10^{-4}

    Using equation 2:

    pK_a=-\log (7.2\times 10^{-4})=3.143

    To calculate the pH of the acidic buffer, the equation for Henderson-Hasselbalch is used:

    pH=pK_a+ \log \frac{\text{[conjugate base]}}{\text{[acid]}}         …….(3)

    Given values:

    [NaNO_2]=\frac{0.0677}{0.150}

    [HNO_2]=\frac{0.04885}{0.150}

    pK_a=3.143

    Putting values in equation 3. we get:

    pH=3.143-\log \frac{(0.0677/0.150)}{(0.04885/0.150)}\\\\pH=3.143-0.142\\\\pH=3.001

    Hence, the pH of the resulting solution will be 3.001

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