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1. A student standrdized a NaOH soln. using 15.00mL of 0.250 M HCl as a primary

ID: 782858 • Letter: 1

Question

1. A student standrdized a NaOH soln. using 15.00mL of 0.250 M HCl as a primary standard. Titration required 42.73 mL of NaOH. Cal the molarity of NaOH soln.


2. A NaOH soln of unknown concentration was standardized against KHP. It took 45.78 mL of NaOH soln to completely titrate 1.256 g of KHP. Cal the molarity of NaOH.


3. Students made the following common errors while standardizing NaOH using KHP as the primary standard. How will each of these errors affect the molarity of NaOH as compared to the correct value? Will it be higher, lower, or no different? Give a reason for each answer.


a. a small amount of KHP is spilled after weighing and while transferring to flask.

b. an air bubble in the buret tip releases in the middle of the titration.

c. the student titrates to a dark pink endpoint.

d. the student adds 150.00mL of water to flask instead of 100.00mL

e the KHP is damp when the student weighs it


Explanation / Answer

ans 1

M1V1 = M2V2

M1*42.73 = 0.25*15

M1 = 0.088 M...........................ANS 1


ans 2

HP- + NaOH ---> NaP- + H2O


so moles of HP- = 1.256/(204.22) = 0.00615


so M1V1 = 0.00615

M1 = 0.00615/0.04578 = 0.134 M....................ANS 2


ans 3

a.

as some amount of KPH is lost the titration would take lesser volume of NaOH

so as M1V1 = moles

moles are constant (as calculated by weighing)

so if V1 is less M1 will be more than actual

or molarity calculated will be more than actual


b.

as the air bubble is present the volume of NaOH will be more than actual required

so molarity calculated will be less than actual


c.

a dark pink endpoint means the volume of base added is more than that at equivalence point

hence as volume is more molarily calculated will be less than actual


d.

as moles of KPH remain constant

molarity will be same


e.

as KPH is damp it means some water is present so actual mass of KPH is less

so moles is less

so volume of NaOH required will be less

so molarity calculated will be more than actual