Question

The pH of 0.50 M HClO is 3.91. Calculate the change in pH when 1.73 g of NaClO (FW = 74.44 g/mol) is added to 40 mL of 0.50 M HClO (FW = 52.46 g/mol). Ignore any changes in volume. The Ka value for HClO is 3.0 x 10-8.

Answer 1

Answer:

The value of change in pH is 3.71

Explanation:

Given:

The pH of HCIO = 3.91

Mole of NaCIO $n_{1} = \frac{1.73}{74.44} = 0.0232$

Moles of HCIO $n_{2} = \frac{0.5 \times 37}{1000} = 0.0185$

Value of Ka $= 3 \times 10^{-8}$

The pH is given by,

  $pH = -pKa + \log \frac{n_{1} }{n_{2} }$

  $pH = -\log(3 \times 10^{-8} ) + \log \frac{0.0232}{0.0185}$

  $pH = 7.62$

Now, the change in pH is given by,

$\Delta pH = 7.62-3.91$

$\Delta pH = 3.71$

Therefore, the value of change in pH is 3.71

Answer 2

The value of change in pH would be "3.71".

pH Value

According to the question,

pH of HCIO = 3.91

Mass of NaClO = 1.73 g

Value of Ka = 3 × 10⁻⁸

Now,

Moles of NaCIO will be:

→ n₁ = $\frac{1.73}{74.44}$

      = 0.0232

Moles of HCIO will be:

→ n₂ = $\frac{0.5\times 37}{1000}$

       = 0.0185

The pH will be:

→ pH = -pKa +log $\frac{n_1}{n_2}$

        = -log (3 × 10⁻⁸) + log $\frac{0.0232}{0.0185}$

        = 7.62

hence,

The change in pH will be:

→ ΔpH = 7.62 - 3.91

           = 3.71

Thus the above answer is appropriate.

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