Answer : The pH of the resulting buffer is, 5.22
Explanation : Given,
[tex]K_a=1.28\times 10^{-5}[/tex]
First we have to calculate the moles of [tex]NaOH\text{ and }HA[/tex]
[tex]\text{Moles of }NaOH=\text{Concentration of }NaOH\times \text{Volume of solution}}=0.160M\times 0.500L=0.08mol[/tex]
and,
[tex]\text{Moles of }HA=\text{Concentration of }HA\times \text{Volume of solution}}=0.200M\times 0.585L=0.117mol[/tex]
The balanced chemical reaction is:
[tex]HA+(aq)+OH^-(aq)\rightarrow H_2O(l)+A^-(aq)[/tex]
Moles of HA left = 0.117 mol - 0.08 mol = 0.037 mol
Moles of [tex]A^-[/tex] = 0.08 mol
The expression used for the calculation of [tex]pK_a[/tex] is,
[tex]pK_a=-\log (K_a)[/tex]
Now put the value of [tex]K_a[/tex] in this expression, we get:
[tex]pK_a=-\log (1.28\times 10^{-5})[/tex]
[tex]pK_a=5-\log (1.28)[/tex]
[tex]pK_a=4.89[/tex]
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
[tex]pH=pK_a+\log \frac{[A^-]}{[HA]}[/tex]
Now put all the given values in this expression, we get:
[tex]pH=4.89+\log (\frac{0.08}{0.037})[/tex]
[tex]pH=5.22[/tex]
Thus, the pH of the resulting buffer is, 5.22
The pH of the resulting buffer is, 5.22
First, we have to calculate the moles of NaOH and HA which is 0.08mol and 0.117mol
Next, after we balance the equation, we now find the moles of A^-= 0.08 mol.
Now, to calculate the pH of buffer:
pH= pKa + log Salt/Acid
= 4.89 + log(0.08/0.037)
=> 5.22
Therefore, the pH of the resulting buffer is, 5.22
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