Question

A chemist titrates 200.0 mL of a 0.7681 M hydrocyanic acid (HCN) solution with 0.5271 M NaOH solution at 25 °C. Calculate the pH at equivalence. The pk of hydrocyanic acid is 9.21. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added. pH- х ?

Answer 1

required at solution Nach + H₂O HON + Naot 200*0.7681 mole HON 10oo HON = 200 0.7681 (M) mL mole HON 0.15 36 for equivalence. is NaOH mL re Leta thation, NaOH 0.5271 x mole mL 0.5271 (M) Nuoh 1000 NaOH x mole 5.271x10 NaOH, I mole with since I mole HON reacts 0.1536 5.271 xlo 0. 1536 5.271810-4 291.406 Produced and will be mole Na en Hence, 0. 1536 molle X1000 Concentration, c' its 0 1536 (200+ 91.406) 0.3126 molla = 0.3126 (1) from weak this that hydrolysis Salt PH hydrolysis acid HON as Condition ean Valence Point) there NacN, Produced and strong base NaOH . such salt is expressed pH = 1/2 Pkw + 1/2 pka + loge PKW = 14 at 빛 11.352 X 11.35 (nounded nounded to 2 decimal 25°c g. + goal # 10p(0-120) L' places) Hence, At equivalence PH 11.35