Help..2007 HSC paper Q21 c

[needsomehelp]

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I don't know how to do this, can anyone help?


10 mL of 0.005 mol L–1 H2SO4 was diluted to 100 mL

What volume of 0.005 mol L–1 KOH is required to neutralise 15 mL of the diluted solution of H2SO4?

[clintmyster]

using the formula c1v1 = c2v2,

C2 = c1v1/v2
= 0.010 x 0.005 / 0.100
= 5 x 10^(-4)

Therefore moles of sulfuric acid = C x V = 5 x 10^(-4) x 0.015 = 7.5 x 10^(-6)

Now 2KOH + H2SO4 --> K2SO4 + 2H2O

There is a 2:1 mole ratio between KOH and H2SO4 Therefore multiple moles of sulfuric acid by 2

now C = n/v therefore V = n/C
so V = 7.5 x 10^(-6) / 0.005 = 0.003 L or 3 mL

[boxhunter91]

Originally Posted by clintmyster

using the formula c1v1 = c2v2,

C2 = c1v1/v2
= 0.010 x 0.005 / 0.100
= 5 x 10^(-4)

Therefore moles of sulfuric acid = C x V = 5 x 10^(-4) x 0.015 = 7.5 x 10^(-6)

Now 2KOH + H2SO4 --> K2SO4 + 2H2O

There is a 2:1 mole ratio between KOH and H2SO4 Therefore multiple moles of sulfuric acid by 2

now C = n/v therefore V = n/C
so V = 7.5 x 10^(-6) / 0.005 = 0.003 L or 3 mL

That looks right.
I remember doing this Question and thinking wth 3mL as an answer.