Question

Calculate the theoretical amounts of 0.1000 M NaOH titrant used to titrate 0.8 and 0.9 g KHP. please show work!

Answer 1

Solution :-

Given 0.8 and 0.9 g KHP

calculate theoretical amounts of the 0.1000 M NaOH

Lets first write the balanced reaction equation

KHP + NaOH ----- > NaKP + H2O

Now lets calculate the moles of the KHP

moles = mass / molar mass

molar mass of KHP = 204.22 g per mol

0.8 g KHP * 1 mol / 204.22 g = 0.0039173 mol KHP

Since mole ratio of the KHP and NaOH is 1 :1

therefore moles of the NaOH needed for the reaction are same as moles of KHP

so moles of NaOH = 0.0039173 mol NaOH

now lets calculate volume of NaOH

Volume in liter = moles / molarity

                    = 0.0039173 mol / 0.10000 mol per L

                    = 0.03917 L

0.03917 L * 1000 ml / 1 L = 39.17 ml NaOH

So volume of NaOH needed to react with 0.8 KHP is 39.17 ml

Now lets calculate for the 0.9 g KHP

0.9 g KHP * 1 mol / 204.22 g = 0.004407 mol KHP

hence moles of NaOH needed are 0.004407 mol

volume of NaOH = 0.004407 mol / 0.1000 mol per L

                         = 0.04407 L

0.04407 L * 1000 ml /1 L = 44.07 ml so we can round it to 44.1 ml

So the amount of NaOH needed to react with 0.9 g KHP is 44.1 ml