Question

Chemistry Help I have a few questions I am working on but I have no idea where to get started, if someone could help me out and explain what you are doing it would be great because I am lost.

These are the questions

A 0.414 g sample of KHP will be titrated with a NaOH solution with an assumed concentration of 0.100 M. Calculate the volume of NaOH that should neutralize the KHP sample.

In a titration of 20.00 mL of 0.1313 M H2SO4, 41.23 mL of an NaOH solution are needed. What is the concentration of the NaOH solution?

25.00 mL of a H2SO4 solution with an unknown concentration was titrated to a phenolphthalein endpoint with 27.05 mL of a 0.1321 M NaOH solution. What is the concentration of the H2SO4 solution?

Answer 1

A 0.414 g sample of KHP will be titrated with a NaOH solution with an assumed concentration of 0.100 M. Calculate the volume of NaOH that should neutralize the KHP sample.

m = 0.414 KHP

MW = 204.22

mol KHP = mass/MW = 0.414/204.22 = 0.002027

NaOH = 0.10 M

find V of NaOH to neutralize KHP

KHP + NaOH --> H2O + NaKP

mol of base = mol of acid

0.002027 = mol of base

M = m/V

V = m/M = 0.002027/0.1 = 0.02027 liter or 20.27 ml

In a titration of 20.00 mL of 0.1313 M H2SO4, 41.23 mL of an NaOH solution are needed. What is the concentration of the NaOH solution?

2 mol of base = 1 mol of acid

mol acid = M*V =20*.1313 = 2.626 H2SO4

then we need 2*2.626 = 5.252 mol of naoh

M = n/L

M = 5.252/41.23 = 0.127M NAOH

25.00 mL of a H2SO4 solution with an unknown concentration was titrated to a phenolphthalein endpoint with 27.05 mL of a 0.1321 M NaOH solution. What is the concentration of the H2SO4 solution?

V = 25 ml of H2SO4

V = 27.07 NaOH M = 0.1321

mol of base = M*V = 27.07*0.1321 = 3.575 mmol of base

we need half of that of acid so 3.575/2 = 1.78

mmol of acid = 1.78

M = m/V

M = 1.78/25 = 0.0712 M