Calculate the pKa of lactic acid, given that when the concentration of free acid is 0.01 M and lactate is 0.087 M and the pH is 4.8.

pH = pKa + log [A-] /[HA] Rearranging the above equation and solving for pKa, we obtainpKa = pH - log [A-] /[HA] = 4.8 - log (0.087/0.01)= 4.8 - log (8.7)= 4.8 - 0.94 pKa = 3.86

Calculate the pH of a mixture of 0.1 M acetic acid and 0.2 M sodium acetate. The pKa of acetic acid is 4.76.

pH = pKa + log [A-] /[HA] = 4.76 + log 0.2/0.1 = 4.76 + 0.301 = 5.06_______

Calculate the ratio of the acetate and acetic acid required in a buffer of pH 5.3.

pH= pKa + log [A-] /[HA] or log [A-] /[HA] = pH - pkalog [A-] /[HA] = 5.3 - 4.76 = 0.54[A-] /[HA] = antilog 0.54 = 3.47

What is the pH of the buffer containing: 0.5 M sodium acetate and 1.0 M acetic acid?

pH = 4.76 + log 0.5/1= 4.76 - 0.301 = 4.46

What will the pH if you dilute 100 ml of 0.5 M NaOH to 1L with distilled water?

# of moles of NaOH = 0.5 M*0.1L = 0.05 mole Total volume after dilution = 1L (it includes 100 ml of 0.5 M NaOH) Now calculate new Molar concentration of the NaOH solution Concentration of diluted NaOH = 0.05/1 = 0.05 M pOH = -log[OH] = -log0.05pOH =-(-1.3) = 1.3 and the pH = 14 - 1.3 = 12.7

What volume of Acetic acid (17.6 N) and what amount (g) of Sodium Acetate would you need to prepare 100 ml of 0.2 M buffer, pH 4.66 (pKa of Acetic acid = 4.76 and MW of Sodium Acetate = 136)

Acetic Acid = 17.6 N Sodium Acetate = 136; pH = 4.66Concentration = 0.2 M; pKa of Acetic acid = 4.76; Volume of buffer = 0.1 L The total number of moles needed = 0.1 L x 0.2 M = 0.02 moles pH = pKa + log [A‑]/ [HA] 4.66 = 4.76 + log [A‑]/ [HA]or log [A‑]/[HA] = 4.66 ‑ 4.76 or log [A]/[HA] = ‑0.1 Now take antilog of ‑0.1 = [A‑]/[HA] = 0.79 ...................... equation 1 Let [A‑] = X, then [HA] = 0.02 ‑ X Plug in the above values in equation 1, X = 0.0158 ‑ 0.79 XX + 0.79X = 0.01581.79 X = 0.0158X = 0.0158/1.79 = 0.00883 moleThus [A‑] = 0.00883 moleAmount of Sodium Acetate (g) = 0.00883 mole x 136 = 1.2 g [HA] = Acetic Acid = 0.02 ‑ 0.00883 = 0.0112 moleAcetic acid = 0.0112/17.6 = 0.000636 L or 0.636 ml

What is the pH of a 10‑8 M HCl solution?

pH = ‑log[H+][H+] = 10‑7 (H2O) + 10‑8 (HCl)= 1 x 10‑7 + 0.1 x 10‑7 = 1.1 x 10‑7pH = ‑log (1.1 x 10‑7)= log 1/1.1 x 10‑7 = log 0.9 x 107 = log 0.9 + log 107 = ‑0.046 + 7 = 6.95

Calculate the pH of a solution prepared by mixing 500 ml of 0.25 M Sodium Acetate with 250 ml 0.1 M HCl.

pH = 5.36

A 0.2 L of 0.05 M solution of dibasic (Na2HPO4) phosphate buffer is mixed with 0.16 L of 0.05 M monobasic (NaH2PO4) phosphate buffer. Calculate pH of the buffer solution

pH = 7.3 when you use pKa =7.2

What are the concentrations of HOAc (Acetic acid) and OAc‑ (Acetate) in a 0.2 M Acetate buffer, pH 5.0? The pKa for HOAc = 4.76

Acetate = 0.127 M and Acetic Acid = 0.073 M