Given , pH = 12, 100 ml + pH = 3, 900 ml .
Fiind the net pH of the mixture.
(answers :- pOH = 4 and pH = 10)
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100 ml of a base of pH = 12. and 900 ml of an acid of pH = 3 are mixed.
In the acid, [ H⁺ ] = 10⁻³ Molar and [OH⁻] = 10⁻¹¹ Molar. So in 0.900 L, there are 0.0009 moles and 0.9*10⁻¹¹ respectively.
pOH of base = 14 - 12 = 2. So concentration of [ OH⁻ ] = 10⁻² Molar
In 100 ml of base there are 0.001 Moles of [ OH⁻ ] & 10⁻¹³ moles of [ H⁺]. [H+] is too small.
When mixed, 0.0009 moles of H⁺ are neutralized by 0.0009 moles of OH⁻ to form water. So 0.0001 moles of OH⁻ remain in the solution.
Total volume of mixed solution = 1.000 L.
So [OH⁻] = 0.0001 M
pOH of solution = - Log₁₀ [ OH⁻ ] = - Log₁₀ 0.0001 = 4
pH = 14 - 4 = 10
In the acid, [ H⁺ ] = 10⁻³ Molar and [OH⁻] = 10⁻¹¹ Molar. So in 0.900 L, there are 0.0009 moles and 0.9*10⁻¹¹ respectively.
pOH of base = 14 - 12 = 2. So concentration of [ OH⁻ ] = 10⁻² Molar
In 100 ml of base there are 0.001 Moles of [ OH⁻ ] & 10⁻¹³ moles of [ H⁺]. [H+] is too small.
When mixed, 0.0009 moles of H⁺ are neutralized by 0.0009 moles of OH⁻ to form water. So 0.0001 moles of OH⁻ remain in the solution.
Total volume of mixed solution = 1.000 L.
So [OH⁻] = 0.0001 M
pOH of solution = - Log₁₀ [ OH⁻ ] = - Log₁₀ 0.0001 = 4
pH = 14 - 4 = 10
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