A solution of sodium hydroxide was standardized by titrating known volumes of th

Question

A solution of sodium hydroxide was standardized by titrating known volumes of the above KHP solution. The resulting data are summarized in the table below:

Concentration of KHP= .0646

What was the concentration of the NaOH in Run 1 of the standardization?

What was the concentration of the NaOH in Run 2 of the standardization?

Concentration of KHP=.0646 RUN 1 RUN 2 Initial KHP buret reading 4.98 0.5 Final KHP buret reading 40.73 33.18 Initial NaOH buret reading 3.14 3.04 Final NaOH buret reading 24.62 22.69

Explanation / Answer

the reaction between KHP and NaOH is given as

KHP + NaOH ---> KNaP + H20

from the above reaction

we can see that

moles of NaOH added = moles of KHP present

now

we know that

moles = conc x volume

so

conc x volume of NaOH = conc x volume of KHP

Mb x Vb = Ma X Va

now

for the first run

we get

volume of KHP (Va)= final reading - initial reading

Va = 40.73 - 4.98

Va = 35.75

given

conc of KHP (Ma) = 0.0646

now

volume of NaOH added (Vb) = 24.62 - 3.14

Vb = 21.48

so

using MbVb = MaVa

we get

21.48x Mb = 35.75 x 0.0646

Mb = 0.107516

so

the conc of NaOH in run1 is 0.107516

now

for the second run

volume of KHP (Va)= final reading - initial reading

Va = 33.18 - 0.5

Va = 32.68

given

conc of KHP (Ma) = 0.0646

now

volume of NaOH added (Vb) = 22.69 - 3.04

Vb = 19.65

so

using MbVb = MaVa

we get

19.65 x Mb = 32.68 x 0.0646

Mb = 0.107437

so

the conc of NaOH in run2 is 0.107437

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