Question

A chemist titrates 220.0 mL of a 0.4581 M hydrocyanic acid (HCN) solution with 0.6417 M KOH 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 KOH solution added. pH = 1 X 5 ?

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Answer 1

At quivalurce point, all Hon has been converted to its conjugate base CN by oddition of kon of now equilibrium mores backward towards he as (N+10 & HeNt on- CN is base, K6= kwa loxto 14 к 6. 1681-10 [Ka = lopker_ 10 -9.21 – 6. 16x16707 K6= 1.62x105 millimales of Hen= molarily x volume (and) 0.4581x 220 -100.78 At quivalence point, millimole of and millimole of base 100.70 - molarity x volume cal, 100-70 = 0.6417 x V (ul) vol of kon= 100.78 0.6417 $157 ml

Total volume: TCN J= 1577220 = 377 ml millimale Total volume 100.18 0.267 m. 377 (N- +H ² HONton- 0.267m 02- tx 0.267-x 1.62x105= [ca x2 0267-N For weak and 0.267>> xt = 0.267 0.267x 1.62x105.4.33x10-6 2.2.08x153 Ton J= x= 2:08x153m . Pon= -log (on-J= -log 2.08x1532.68 - 14-pon: 14-2-68 = 11. 32