Answer:
The value of change in pH is 3.71
Explanation:
Given:
The pH of HCIO = 3.91
Mole of NaCIO [tex]n_{1} = \frac{1.73}{74.44} = 0.0232[/tex]
Moles of HCIO [tex]n_{2} = \frac{0.5 \times 37}{1000} = 0.0185[/tex]
Value of Ka [tex]= 3 \times 10^{-8}[/tex]
The pH is given by,
[tex]pH = -pKa + \log \frac{n_{1} }{n_{2} }[/tex]
[tex]pH = -\log(3 \times 10^{-8} ) + \log \frac{0.0232}{0.0185}[/tex]
[tex]pH = 7.62[/tex]
Now, the change in pH is given by,
[tex]\Delta pH = 7.62-3.91[/tex]
[tex]\Delta pH = 3.71[/tex]
Therefore, the value of change in pH is 3.71
The value of change in pH would be "3.71".
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 = [tex]\frac{1.73}{74.44}[/tex]
= 0.0232
Moles of HCIO will be:
→ n₂ = [tex]\frac{0.5\times 37}{1000}[/tex]
= 0.0185
The pH will be:
→ pH = -pKa +log [tex]\frac{n_1}{n_2}[/tex]
= -log (3 × 10⁻⁸) + log [tex]\frac{0.0232}{0.0185}[/tex]
= 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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