Calculate the pH of a buffer that is 0.190 M in NaHCO3 and 0.265 M in Na2CO3. Ca

Question

Calculate the pH of a buffer that is 0.190 M in NaHCO3 and 0.265 M in Na2CO3.

Calculate the pH of a solution formed by mixing 65 mL of 0.35 M NaHCO3 with 75 mL of 0.18 M Na2CO3.

Explanation / Answer

(HCO3)-1 --> H+ & (CO3)-2 [0.190M] ....... [H+] .....[0.165M] Ka for (HCO3)-1 5.6e-11 = [H+] [0.165] / [0.190] [H+] = 6.4484e-11 pH+ = 1.032 your answer = 1.032 ------------------- 2) pH = pKa + log([A-]/[HA]) (Henderson-Hasselbalch Eqn.) C1V1 = C2V2, or C2 = C1V1/V2. Volume after mixing = V2 = 65 mL + 75 mL = 140 mL. [A-] = [CO3--] = 0.18 M * 75 mL / 140 mL = 0.0964 M. [AH] = [HCO3-] = 0.35 M * 65 mL / 140 mL = 0.1625 M. Substituting, pH = -log(5.6x10^-11) + log(0.0964 / 0.1625) = 10.025.

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