Question

How Many ml of 0.4M NaOH would exactly neutralize a mixture formed from 20ml of 0.1M KOH and 30 Ml of 0.1M h2so4?

Answer 1

first of all we need to find out what happened when we added the potassium hydroxide to the sulphuric acid:

recall acid + base -> salt + water
\[H_2 SO_4 + KOH \rightarrow K_2 SO_4 + H_2O\]

now to balance:
\[H_2 SO_4 + 2KOH \rightarrow K_2 SO_4 + 2H_2O\]

so from this we see that 1 mol of sulphuric acid reacts with 2 moles of potassium hydroxide

now we should find the amounts we added together:

remember: amount of substance (mol)= Concentration x Volume

\[\text{for KOH : } 0.1mol/dm^3 \times 0.02 dm^3 = 0.002 mol\]
\[\text{for } H_2SO_4 \text{ : } 0.1 mol/dm^3 \times 0.03 dm^3 = 0.003 mol\]

so because there is less KOH (we would need 2 x 0.003 for the acid to be used up)
it is the limiting factor

i have to go :( i hope someone else can continue

Answer 2

well what are you unclear about? eigenschmeigen told you what to do...

Answer 3

so 0.002 mol of KOH reacts with 0.001 mol sulphuric acid, leaving 0.002 unreacted

now we need reaction of NaOH with remaining sulphuric

\[H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O\] once again, we have 1:2

0.002 mol of sulphuric would need 0.004 mol of NaOH , so now we need to know how much 0.4M naOH is 0.004 mol

Answer 4

\[0.4 mol/dm^3 = \frac{0.004mol}{V}\]

so rearrange, V = 0.01 dm^3 = 10 ml