4.235 is the pH of a solution that is 0.175M benzoic acid and 0.190M sodium benzoate, a salt whose anion is the conjugate base of benzoic acid.
According to Henderson Hasslbalch Equation to calculate the pH of a solution is expressed as
[tex]pH = pK_{a} + \log \frac{[\text{Congujate base}]}{[\text{Weak acid}]}[/tex]
Benzoic acid (C₆H₅COOH) and its conjugate base is Benzoate (C₆H₅COO⁻) are considered as acidic buffer.
Kₐ for benzoic acid = 6.3 × 10⁻⁵
So, pKₐ = -log Kₐ
= - log [6.3 × 10⁻⁵]
= 4.2
Now put the value in above expression we get
[tex]pH = pK_{a} + \log \frac{[\text{Congujate base}]}{[\text{Weak acid}]}[/tex]
[tex]pH = pK_{a} + \log \frac{C_6H_5COONa}{C_6H_5COOH}[/tex]
[tex]= 4.2 + \log \frac{0.190}{0.175}[/tex]
= 4.2 + 0.035
= 4.235
Thus from the above conclusion we can say that 4.235 is the pH of a solution that is 0.175M benzoic acid and 0.190M sodium benzoate, a salt whose anion is the conjugate base of benzoic acid.
Learn more about the Henderson Hasslbalch Equation here: https://brainly.com/question/9255613
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