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April 18, 2014

April 18, 2014

Posted by **jo** on Thursday, November 29, 2012 at 8:36pm.

I have been trying to figure out this titration equation all day and for some reason I am stumped.

Here is the question in full:

Potassium hydrogen phthalate, KHP, can be obtained in high purity and is used to determine the concentrations of solutions of strong bases. Strong bases react wiht the hydrogen phthalate ion as follows:

HP- (aq) + OH- = H2O (l) +p2- (aq)

The molar mass of KHP is 204.2 g/mol and Ka for the HP- ion is 3.1 x 10^-6.

The Questions are: (a) If a titration experiment begins with 0.4885 g of KHP and has a final volume of 100ml, what is the ph of the equivalence point?

(b) If the titration required 38.55ml of NaOH solution to reach the end point, what is the concentration of the NaOH solution?

Thanks for anyone's help and direction with this.

- Chemistry -
**DrBob222**, Thursday, November 29, 2012 at 9:29pma.

mol KHP = 0.4885g/molar mass = ?

M = mols/0.1 L = approximately 0.024 but you need to do it more accurately.

At the equivalence point the pH is determined by the hydrolysis of the P^2- ion.

.........P^2- + HOH ==> HP^- + OH^-

I.......0.024M...........0.......0

C..........-x...........x........x

E.......0.024-x.........x........x

Kb for P^2- = (Kw/k2 for H2P) = (x)(x)/(0.024-x)

Substitute all of the numbers and solve for x = OH^-, then convert to pH.

b.

mols KHP = 0.4485/molar mass KHP

mols NaOH = mols KHP (the equation is 1:1)

M NaOH = mols NaOH/L NaOH. You know mols NaOH and L NaOH solve for M NaOH.

- Chemistry -
**jo**, Friday, November 30, 2012 at 1:28amTHANK YOU!

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