Calculate the pH of a solution made by mixing 50.0 mL of 0.100 M NaCN with 11.82

Question

Calculate the pH of a solution made by mixing 50.0 mL of 0.100 M NaCN with 11.82 mL of 0.438 M HClO4 . pKa for HCN = 9.21 Please show ALL work. If work is not shown, i will not rate lifesaver

Explanation / Answer

HClO4 is a very strong acid, so it will react completely with NaCN to produce the acid form: NaCN + HClO4 ----> HCN + NaClO4 A) 50.00 mL * 0.100 M NaCN = 5 mmol NaCN 4.20 mL * 0.438 M HClO4 = 1.84 mmol HClO4 Since the reaction essentially goes to completion, you will have: 5-1.84 = 3.16 mmol NaCN 1.84-1.84 = 0 mmol HClO4 0+1.84 = 1.84 mmol HCN NaCN wil be in equilibrium by dissociating to form its acid form in water. ....NaCN + H2O HCN + OH- I. . .3.16 . . . . . . . . . . .1.84. . .0. . . C. . .-x . . . . . . . . . . . . . .+x . . .+x E. . .3.16-x . . . . . . . . .1.84+x. . .x . Kb = [HCN][OH-] / [NaCN]; kb = 1E-14/ka = 1.6E-5 1.6E-5 = (1.84+x)(x) / (3.16-x) Assume x is going to be small, and check assumption by methods of successive approximation: x1 = 2.75E-5 x2 = 9.32E-6 x3 = 9.32E-6 x3 = [OH-] = 9.32E-6 = pOH = 5.03 pH = 8.97 You do the same for the 11.82 mL of 0.438 M HClO4. At the equivalence point, you will have equal moles (or mmoles) of HClO4 and HCN. To find out how many mL that is, use the equation, C1V1 = C2V2. Hope that helps.

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