Question

A volume of 500.0 mL of 0.100 M NaOH is added to 545 mL of 0.200 M weak acid (K, 7.44 x 10). What is the pH of the resulting buffer? HA(aq) + OH (aq) H,O)+A (aq) 4.208 PH

Answer 1

No. of moles of NaOH in 500.0 mL of 0.100 M NaOH = 500.0x 0.100 molly 1000 0.0500 mol Again, no. of moles of weak acid, HA in 545 mL of 0.200 M HA = 545 x x 1000 0. 200 mol/x 0.05 mol = 0.10g mol Here the given reaction - .HA (ar) + OH (ar) H20(e) + Á (aq) 1 mol 1 mot 1 mol 0.05 mol 0.05 mol Thus, the no. of moles of unreacted HA = (0.109 - 0.05) mol = 0.05g mol No. of moles of A in the reaction medium = 0.05 mol

Now, total volume of the solution = (500 + 545) mL = 1045 mL = 1.045 L .. Molarity of unreacted HÀ, [HA] = 0.059 mol 1.045 L = 0.0564 M Molarity of a [A-] = 0.05 mol 1.045 L = 0.0478 M Given, Ka = 7.44 x 10-5 - log ka = -log (7.44 x 10-5) a pka = 4.128 Now, pH of the butter solution can be given by the Henderson equation as below -

pH = pka + log [A-] THA = 4.128 + log 0.0478 0.0564 = 4.056 ~ 4.06 [three significant figures] Thus the pH of the buffer solution is - pH = (4.06 Aus
Here the given answer is in three significant figures........

If the question demand 4 significant figures then try the value of pH = 4.056