How much 0.985 m koh is needed to to titrate 50.00 ml of 2.32 m hno3?

The answer to your question is: 118 ml of KOH

Explanation:

Data

volume KOH = ?

[KOH] = 0.985 M

volume HNO3 = 50 ml = 0.05 l

[HNO3] = 2.32 M

Reaction

                KOH + HNO3 ⇒  KNO3 + H2O

Formula

Molarity = moles / volume

Process

moles of HNO3 = 2.32 x 0.05 l

                          = 0.116

From the equation we know that the proportion KOH : HNO3 = 1 : 1

then,                    

                        1 mol of KOH  1 mol of HNO3

                        x                        0.116 mol of HNO3

                        x = (0.116 x 1)/ 1

                        x = 0.116 moles of KOH

Now, calculate volume of KOH

Volume of KOH = moles of KOH / molarity

Volume of KOH = 0.116 / 0.985

                          = 0.118 l of KOH or 118 ml

the maximum mass of nh3 is nh3 in grams