Question 59 of 69intermediate🔧 ApplyNumerical3 marks

The ionization constant of nitrous acid is 4.5×10−44.5 \times 10^{-4}. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Correct Answer

pH = 7.97, Degree of hydrolysis = 2.35×10−52.35 \times 10^{-5}

Exercise: EXERCISES | Q: 6.61 | (Page 46)
For More Understanding

Explanation

Sodium nitrite is a salt of a weak acid (nitrous acid) and a strong base (NaOH). The nitrite ion undergoes hydrolysis in water to produce OH- ions, making the solution basic. We use the hydrolysis constant to find the degree of hydrolysis and the concentration of OH- ions, which then allows us to calculate the pH.

Solution Steps

  1. Step 1: Identify the salt type and write the hydrolysis reaction Sodium nitrite (NaNO2NaNO_2) is a salt of a weak acid (HNO2HNO_2) and a strong base (NaOHNaOH). The NO2−NO_2^- ion hydrolyzes in water:

    NO2−+H2O⇌HNO2+OH−NO_2^- + H_2O \rightleftharpoons HNO_2 + OH^-

  2. Step 2: Calculate the hydrolysis constant (KhK_h)

    Kh=KwKaK_h = \frac{K_w}{K_a}

    Given Ka=4.5×10−4K_a = 4.5 \times 10^{-4} and Kw=1×10−14K_w = 1 \times 10^{-14}

    Kh=1×10−144.5×10−4=2.22×10−11K_h = \frac{1 \times 10^{-14}}{4.5 \times 10^{-4}} = 2.22 \times 10^{-11}

  3. Step 3: Calculate the degree of hydrolysis (hh) For a salt of weak acid and strong base, h=Khch = \sqrt{\frac{K_h}{c}} Given concentration c=0.04c = 0.04 M

    h=2.22×10−110.04=55.5×10−11=2.35×10−5h = \sqrt{\frac{2.22 \times 10^{-11}}{0.04}} = \sqrt{55.5 \times 10^{-11}} = 2.35 \times 10^{-5}

  4. Step 4: Calculate the concentration of OH−OH^- ions [OH−]=c×h=0.04×2.35×10−5=9.4×10−7[OH^-] = c \times h = 0.04 \times 2.35 \times 10^{-5} = 9.4 \times 10^{-7} M

  5. Step 5: Calculate the pH of the solution

    pOH=−log⁡[OH−]=−log⁡(9.4×10−7)=6.03pOH = -\log[OH^-] = -\log(9.4 \times 10^{-7}) = 6.03 pH=14−pOH=14−6.03=7.97pH = 14 - pOH = 14 - 6.03 = 7.97
PreviousNext