Question #92988
A 25.00 mL aliquot of KHP prepared in (a) required 23.85 mL of NaOH solution to reach
the end point during titration
(i) Write a balanced equation of the reaction
(ii) What is the exact concentration in moles/Liter of your NaOH solution?
(iii) Which one between KHP and NaOH is the primary standard and why?
(iv) Which indicator is suitable for a weak-acid strong-base titration and why?
1
Expert's answer
2019-08-21T04:54:39-0400

(I)

KHP+NaOH=NaKP+H2OKHP + NaOH = NaKP + H_2O

In ionic form:


H++OH=H2OH^+ + OH^- = H_2O

(II)

CKHPVKHP=CNaOHVNaOHC_{KHP} V_{KHP} = C_{NaOH} V_{NaOH}


CNaOH=CKHPVKHPVNaOHC_{NaOH} = {C_{KHP} V_{KHP}\over V_{NaOH}}

Accordingly, for an accurate calculation of the molar concentration of NaOH, it is necessary to know the exact concentration of KHP.  Suppose that the concentration of K is 0.1 mol/L, then:

CNaOH=0.12523.85=0.105.C_{NaOH} = {0.1 * 25 \over 23.85} = 0.105.


Unfortunately, the condition of the assignment does not indicate the mass of dissolved KHP or its exact concentration (insufficient data for a complete solution). The above calculation is an example of a solution.


(III) KНР provides an excellent primary standard for a number of reasons. So, it is not hygroscopic, has high purity, is soluble, non-toxic, stable, and affordable.


Also, this reaction is very convenient in terms of molar ratios (1: 1).


(IV) It is convenient to use phenolphthalein as an indicator.

First of all, phenolphthalein is a one-color indicator (transparent in acidic or neutral environment and raspberry in alkaline). The presence of at least one "extra drop" of NaOH will immediately affect the color - the solution will instantly become raspberry.


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