Answer to Question #120864 in Inorganic Chemistry for Anais

Question #120864
Given that the Ka of benzoic acid is 6.3 x10-5, calculate the pH when 0.080M sodium hydroxide is added to 10.0 ml of 0.052M benzoic acid at the equivalence point
1
Expert's answer
2020-06-08T15:42:02-0400

Here in the question, "K_a(C_6H_5COOH)=6.3 \\times 10^{-5}" , 0.08 M of NaOH is aded to 10


mL of 0.052 M "C_6H_5COOH"


The reaction between them is taking place as


"C_6H_5COOH + NaOH \\longrightarrow C_6H_5COONa + H_2O"

and,

"C_6H_5COOH \\longrightarrow C_6H_5COO^- + H^+"


"K_a= \\frac{[C_6H_5COO^-][H^+]}{[C_6H_5COOH]}"


"6.3\\times 10^{-5}=\\frac{[0.08 +x][x]}{[0.05-x]}"


x= "3.9\\times 10^{-5}" = "[H^+]"

now, pH =-log "[H^+]"


pH= 4.408



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