What is the pH of a buffer that was prepared by adding 3.96 g of sodium benzoate, #NaC_7H_5O_2#, to 1.00 L of 0.0100 M benzoic acid, #HC_7H_5O_2#?

Question

Assume that there is no change in volume. The #K_a# for benzoic acid is #6.3 * 10^-5#.

Mateo Aguilar · Accepted Answer

The #"pH"# of the buffer is 4.64.

Let's represent the benzoate ion, #"C"_7"H"_5"O"_2^"-"#, as #"A"^"-"#. After that, the benzoic acid dissociation equation becomes #"HA" + "H"_2"O" ⇌ H_3"O"^+ + "A"^"-"#; #K_"a" = 6.3 ×10^"-5"# The typical formula for calculating the #"pH"# of a buffer is the Henderson-Hasselbalch equation: #color(blue)(|bar(ul("pH"= "p"K_"a" + log (("[A"^"-""]")/"[HA]"))|)# It is our responsibility to compute the values that go into the formula. (a) #"p"K_"a" = "-"logK_"a" = "-"log(6.3 × 10^"-5") = 4.20# (b) #[A^"-"] = ["C"_7"H"_5"O"_2^"-"] = ["NaC"_7"H"_5"O"_2]# #3.96 color(red)(cancel(color(black)("g NaC"_7"H"_5"O"_2))) × ("1 mol NaC"_7"H"_5"O"_2)/( 144.10 color(red)(cancel(color(black)("g NaC"_7"H"_5"O"_2)))) = "0.027 48 mol NaC"_7"H"_5"O"_2# ∴ #[A^"-"] = "0.027 48 mol"/"1 L" = "0.027 48 mol/L"# (c) #"[HA] = 0.0100 mol/L"# These values are now entered into the Henderson-Hasselbalch formula. #"pH"= "p"K_"a" + log (("[A"^"-""]")/"[HA]") = 4.20 + log(("0.027 48" color(red)(cancel(color(black)("mol/L"))))/ (0.0100 color(red)(cancel(color(black)("mol/L"))))) = 4.20 + log(2.748) = 4.20 + 0.439 = 4.64"# The #"pH"# of the buffer is 4.64.

Cooper Anderson · Answer

undefined The pH is 4.18.