In a titration of 47.41 ml of 0.3764 m ammonia with 0.3838 m aqueous nitric acid, what is the ph of the solution when 47.41 ml + 10.00 ml of the acid have been added?

The pH of the solution is 1.136

Explanation:

To calculate the moles from molarity, we use the equation:

Molarity of ammonia = 0.3764 M

Volume of ammonia = 47.41 mL = 0.04741 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

Molarity of nitric acid = 0.3838 M

Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L

Putting values in above equation, we get:

After the completion of reaction, amount of nitric acid remained = 0.022 - 0.0178 = 0.0042 mol

For the reaction of ammonia with nitric acid, the equation follows:

At             0.0178   0.022

Completion        0     0.0042        0.0178

As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.

To calculate the pH of the reaction, we use the equation:

where,

Putting values in above equation, we get:

Hence, the pH of the solution is 1.136

i believe the answer is three water molecules