158.4:1 159 0.050 158:1158: 11158.4:1|158.49 Match 0.67/1 159 0.050310.0503 mol/

Question

158.4:1 159 0.050 158:1158: 11158.4:1|158.49 Match 0.67/1 159 0.050310.0503 mol/0.0503mol/LI Q: What is the molar ratio of a base:salt buffered solution with a pH of 7.8 and a pKb of 8.4? If the concentration of base (HB+) is 0.05 mol/L, what is the total molarity of the solution? guided questions: -what is the molar ratio of the solution?1 -How many total parts (base + salt) are in the solution? 2_parts (3 sig fig)-What is the total molarity of the solution?-3_mol/L (3 sig fig) B1:158:11158 11158.4:1]158.49:11 B2: 159 B3:0.050310.0503 mol/0.0503mol/L

Explanation / Answer

For the given buffer problem

we would use Hendersen-Hasselbalck equation,

pH = pKa + log(base/salt)

pKb = 8.4

pH = 7.8

So,

7.8 = (14 - 8.4) + log(base/salt)

[with pKa = 14 - pKb]

molar ratio of base : salt = 158.49 : 1

Total molarity of solution = 158.49 + 1 = 159.49 = 159 (with 3 sig fig)

When [BH+] concentration = 0.05 mol/L

[B] = (1/159) x 0.05 = 0.0003 mol/L

Total molarity of solution = [BH+] + [B]

                                        = 0.05 + 0.0003

                                        = 0.0503 mol/L (with 3 sig fig)

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