# What is the molarity of the NaOH solution?

A student performing an experiment titrates 1.390 g of KHP with an NaOH solution of unknown concentration. The initial buret reading is 0.73 mL and the final buret reading is 22.38 mL. What is the molarity of the NaOH solution?

Please help with easy and clear calculations.

Answer:
They react 1:1
M.Wt KHP= 204.22 so 1.39g=1.39/204.22=0.006806mole...
0.006806 moles/21.65ml is sames as=0.006806x1000/21.65 = 0.3144 M per litre
simply:
[NaOH] = 1000 x wt / titre x 204.22
= 1390 / (22.38-.73 ) x 204.22
= 0.3144 M
Here is your balanced equation:

KHP(aq) + NaOH(aq) >> KNaP + H20
They react in a 1 to 1 ratio

1.390g KHP x ..1 mole KHP..=
......204.22... g KHP

6.806x10-3 moles of KHP. Since you have a 1 to 1 ratio
you have 6.806x10-3 moles of NaOH

Now simply dividethe number of moles by the volume titrated in liters.

6.806x10-3/0.02165L = 0.3144M

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