{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Buffer for 4.5 pH

# Buffer for 4.5 pH - 4.5 = 4.745 log10(0.10[NaC2H3O2 0.245 =...

This preview shows page 1. Sign up to view the full content.

Buffer question involving Ka and pH? Acetic acid has a Ka value of 1.8e-5. How many grams of NaC2H3O2 would have to be added to 100 ml of .10 M HC2H3O2 to prepare a buffer with a pH of 4.5? Best Answer - Chosen by Voters Using Henderson Hesselbach equation: pH = pKa - log10([acid]/[conjugate base]) pH = pKa - log10([HC2H3O2]/[NaC2H3O2]) ______________________________________. .. 1st we should find the pKa Ka = 1.8e-5 pKa = -log(Ka) = -log(1.8e-5) = 4.745 ______________________________________. .. We have to find the concentration of the conjugate base [NaC2H3O2] = ???? pH = 4.5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 4.5 = 4.745 - log10(0.10/[NaC2H3O2]) 0.245 = log10(0.10/[NaC2H3O2]) 0.245 = log10(0.10) – log10([NaC2H3O2]) log10([NaC2H3O2]) = -1 – 0.245 = - 1.245 [NaC2H3O2] = 10^(-1.245) = 0.056885 ______________________________________. .. n = C*V à moles = concentration (mol/L) * Volume (L) n = = 0.056885 mol/L * 0.1L = 5.6885 x 10-3 ______________________________________. .. m = n*M à mass = mole * molecular weight molecular weight of NaC2H3O2 = 82.024 …. . (from your periodic table) therefore m = 82.024 x 5.6885 x 10-3 = 0.4666 g ANSWER = 0.4666 g...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online