Question

Calculate the pH of the solution that results from each of the following mixtures.

A) 55.0 mL of 0.15 molL−1 HCOOH (Ka=1.8×10−4) with 80.0 mL of 0.13 molL−1 HCOONa.
B) 135.0 mL of 0.13 molL−1 NH3 (Kb=1.76×10−5) with 240.0 mL of 0.13 molL−1 NH4Cl

Answer 1

155.0 mL a 0.15 molt HCOOH = (0.15 x 55.0) mol HCOOH = 8.25*10-3 mol HCOOH 30.0 me af 0.13 malah Helona = (0.13 x 80.0) mel Hero Na 0.0104 mol HCOONa " according to the Henderson equation, pH = pka + log [salt] on ka = 1.8810-4 ple = -by na = -log (1.8 x 10 %) = 3.74 and Tacid ] pH = 3.74 + Log 0.0104 8.25x 10-3 = 3.84 y the solution will be 3.84 Hence, the pH 13510 ml of 0.13 moltt NH3 = (0.18 x 135.0) met NH3 = 0.01755 mol Nt3 240.0 mL f 0.13 mol Lt Nity U = 10.13 x 240.0) mol NHGU Ισμο = 0.0312 Now, ml Nige [salt] according to the Henderson equation, он - р o + log - ел [base] Geiven, Ko=1.76x10-5 ple = -log ks = – log (1.76 x 10-5) = 4.75 POH = 4.75 + log 0.0312 = 4.999 ²5 pH = 14 - POH = 14-5=9 Hence, the pH of the solution will be 9