Chemistry- Help!!!!!!?

Question

A sample of 20.0 milliliters of a 0.100-molar HCN solution is titrated with a 0.150-molar NaOHsolution. (Ka HCN=6.2 x 10^-10).
a)What volume of NaOh is used in the titration in order to reach the equivalence point?
b)What is the molar concentration of CN- at the equivalence point?

Answer 1

Lancenigo di Villorba (TV), Italy

Your experiment interests an Acid-Base Equilibrium.
You execute a Titrimetric Determination where a well-defined Hydrocyanic Acid's Analyte undergies Neutralisation by adding drop by drops the Titrimetric Reactive as Sodium Hydroxide's solution.
The overwritten assumpts agree the SODIUM HYDROXIDE LIKE A STRONG BASE ABLE TO NEUTRALISE HYDROCYANIC ACID

HCN(aq) + NaOH(aq) <---> NaCN(aq) + H2O(aq)

QUESTION a)
The Titrimetric End-Point overcomes at the Molar Amount's Correspondance, as it is in the following calculations :

(HCN's Moles) = (HCN's Volume) * (HCN's Molarity)
(HCN's Moles) = 20.0E-3 * 0.100 = 2.0E-3 mol
(HCN's Moles) = (NaOH's Moles)
(NaOH's Moles) = (NaOH's Molarity) * (NaOH's Volume)
(NaOH's Volume) = (NaOH's Moles) / (NaOH's Molarity)
(NaOH's Volume) = 2.0E-3 / 0.150 = 1.3E-2 liter

QUESTION b)
At the Titrimetric End-Point's Conditions, I assumed that Hydrocyanic Acid became Sodium Cyanide.
A more precise consideration involves the Chemical Equilibrium of WEAK ACID's AQUEOUS SOLUTIONs, as it does HYDROCYANIC ACID.
In effects, any aqueous solutions of Hydrocyanic Acid obey to

HCN(aq) <---> H+(aq) + CN-(aq)
6.2E-10 = Ka = |CN-| * |H+| / |HCN|

while Water undergoes its own Ionic Dissociation Phenomena

H2O(aq) <---> H+(aq) + OH-(aq)
1.0E-14 = Kw = |H+| * |OH-|

On the other hands, I need to warrant the Molar Amount's Balancements (e.g. Cyanide Radicals cannot form or sink themselves)

|HCN|0 = (Total Cyanide Radicals) =
= (HCN's Moles) / (Final Volume) =
= 2.0E-3 / (20.0E-3 + 33.3E-3) = 6.0E-2 M
|HCN|0 = |HCN| + |CN-|

as it is for the related Electrical Charge One (e.g. aqueous solutions are electrically neutral ones)

|H+| + |Na+| = |OH-| + |CN-|
|H+| + |HCN|0 = |OH-| + |CN-|

By an analytical solver I found the numerical solutions, e.g.g the following ones

|OH-| = 9.8E-4 M
|H+| = 1.0E-11 M hence pH = 11.0
|CN-| = 6.0E-2 M
|HCN| = 9.8E-4 M.

I hope this helps you.

Answer 2

Volume of NaOH = 20 x 0.1/0.15 = 13.3cm3.

Moles CN- = 2.0 x 10-3. Total volume = 33.3cm3.

Now you can work out the molarity.

Answer 3

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