How much of 0.5M H2SO4 is needed to neutralize 50.0 ml of 0.2M KOH?

Question

thanks

Answer 1

(50 mL)(L/ 1000 mL)(.2 mol/ L)(L/ 0.5 mol)(1000 mL/L) = 20 mL of H2SO4

You have to relate the moles of analyte (KOH) to the moles of the titrant (H2SO4). We can perform this kind of calculation with the knowledge that at the endpoint the titrant will have stoichiometrically reacted with all of the analyte so that the moles of titrant equals the initial moles of analyte.

Answer 2

How much KOH do you have?

50 mL of 0.2 mole/L = 10 mM (milli Mole)

The reaction is 2KOH + H2SO4 -> K2SO4 + 2H20

So 1 mole of H2SO4 neutralizes 2 moles of KOH

Therefor you need 5 mM of H2SO4.

5 mM = (x mL of solution) * (0.5 mole / L)

Solve for x:

x = 10 mL of H2SO4

Answer 3

When you do a problem of this type, there is a specific method that you must use so that you avoid getting the incorrect answer. As above, you have 2 different answers, so which is correct, and how do you avoid this problem. If yoiu follow this method correctly, you will always be correct, no matter whether you deal with a monoprotic, diprotic or triprotic acid (or base for that matter)

1. Write a balanced equation:

H2SO4 + 2KOH → K2SO4 + 2H2O

2 Work with the reactants only - ignore the products.

1mole H2SO4 reacts with 2 mol KOH

3 Now substitute into this equation, where:

nA = moles of acid (from above balanced equation

nB = moles of base

CA = concentration of acid

VA = volume of acid

CB = concentration of base

VB = volume of base

nA/nB  = CA*VA/CB*VB

nA    CA*VA

---- = -------------

nB    CB * VB

1......0.5  *  VA

--- = ---------------

2 .......2  *  50   (sorry this format does not allow for the easy writing of equations. ignore the dots)

VA * 0.5*2  =  1*0.2*50

VA = 10/1

Va = 10ml

Answer The volume required is 10 ml of H2SO4

Just work like this and you will always be OK