# It was found that 25.20mL of an HNO3 solution is needed to react completely with 300mL of LiOH solution that has a pH of 12.05. What is the molarity of HNO3 solution?

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Given

Ph of the solution = 12.05.

We know PH + POH = 14

12.05 + POH = 14

POH = 14 - 12.05

POH = 1.95

We know POH = -[log OH-]

1.95 = -log OH-

Applying Anti log on both sides

OH- = 0.011

OH- ions is obtained from LiOH

LiOH -------> Li+ + OH-

Therefore 0.011 is the concentration of LiOH, nothing but molarity of LiOH

We have ....

LiOH + HNO3-----> LiNO3 + H2O

Molarity of LiOH (M1) = 0.011

Volume of LiOH (V1) = 300ml = 0.3 lts

Molarity of HNO3 (M2) = ?

Volume of HNO3 (V2) = 25.20 ml = 0.0252 lts

By the formula of Dilution...

M1V1/n1 = M2V2/n2

(0.011 * 0.3)/1 = (M2 * 0.0252)/1

0.0033 = M2 * 0.0252

M2 = 0.13

Therfore 0.13 is the concentration of HNO3.

Now pH of LiOH is 12.05.

Now we know that

pH + pOH = 14

so

pOH = 14-pH = 14-12.05 = 1.95

And

pOH = -log[OH]

[OH] = 10^-pOH = 10^(-1.95) = 0.011

So the concentration of OH is 0.011 M

Now LiOH will disassociate into

LiOH =====> Li (+) + OH(-)

here the molar ratio is 1 LiOH : 1 (OH-)

so the concentration of LIOH is 0.011 M

then

M1 = molarity of LiOH = 0.011 M : V1 = volume of LiOH = 300 mL

M2 = molarity of HNO3 = ? : V2 = volume of HNO3 = 25.20 mL

By titration formula,

M1V1=M2V2

M2 = M1V1/V2 = (0.011*300)/25.20 = 0.13 M

So the molarity(concentration) of HNO3 is 0.13 M