You have 200 ml of 0.1M HEPES buffer, pH 7.5. You add 2 ml of 0.4M...

You have 200 ml of 0.1M HEPES buffer, pH 7.5. You add 2 ml of 0.4M HCl. The pKa of HEPES is 7.5.

a) what will the resulting pH be?

b) what would the pH be if no HEPES were present (that is, you mixed water and HCl)?

Solutions

Expert Solution

(a): Given the total initial concentration of the buffer solution = 0.1 M

Since pH = pKa, [salt] = [acid] = 0.1 M / 2 = 0.05 M

Hence initial moles of HEPES salt = MxV = 0.05 mol/L x 0.200 L = 0.01 mol

initial moles of HEPES acid = MxV = 0.05 mol/L x 0.200 L = 0.01 mol

Volume of 0.4 M HCl added = 2 mL = 0.002 L

Hence moles of HCl = MxV = 0.4 mol/L x 0.002 L = 0.0008 mol

------------ HEPES salt + HCl ---------- > HEPES Acid + Cl-

init.mol: 0.01 mol, ------ 0.0008 mol, --- 0.01 mol

change: - 0.0008 mol, - 0.0008 mol, + 0.0008 mol

eqm.mol: 0.0092 mol, -- 0 mol, ---------- 0.0108 mol

Now pH can be calculated from Hendersen equation

pH = pKa + log[HEPES salt] / [HEPEs acid]

=> pH = 7.5 + log(0.0092 mol / 0.0108 mol) = 7.43 (answer)

(b) If there were no buffer,

Volume of 0.4 M HCl added = 2 mL = 0.002 L

Hence moles of HCl = MxV = 0.4 mol/L x 0.002 L = 0.0008 mol

If initial volume of water is 200mL (=0.200L), total volume of solution, Vt = 0.200 L + 0.002L = 0.202 L

[HCl] = [H+] = 0.0008 mol / 0.202 L = 0.00396 M

=> pH = - log[H+] = - log(0.00396M) = 2.40 (answer)

queen_honey_blossom answered 1 year ago

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