A solution of NaOH was standardized by the procedure described in this laborator

Question

A solution of NaOH was standardized by the procedure described in this laboratory. An average of 42.12mL of the base was required to titrate 15.00mL of 0.2878M HCI standard to its end point. The base was then used to determine the concentration of an unknown citric acid solution. Citric acid has three acidic protons and can be abbreviated H,Cit. Calculate the concentration of the unknown HsCit solution if 10.00mL of the solution required an average of 46.87mL of the standardized base. (Assume that all three protons in H,Cit are titrated in the experiment.) 2.

Explanation / Answer

1) standarization

M1V1 = M2V2

42.12 mL x M base = 15.0 mL x 0.2878M

Thus molarity of base = 0.1025 M

2) calculation of concentration of citric acid

The neutralization equation is

H3Cit + 3 NaOH -------------> Cit Na3 +   3H2O

Then  

M1V1 /n1 = M2V2/n2

Mcit x 10mL /1 = 0.1025 x 46.87 /3

Thus molarity of citric acid = 0.160 M

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