Hi! Please can you explain how to do this question?
'What volume of 0.005 mol L-1 KOH is required to neutralise 15mL of 0.0005 mol L-1 H2SO4?'
The equation is H2SO4 + 2KOH -> K2SO4 + 2H2O so isn't the number of moles of H+ already equal to the number of moles of OH-? In the answers they found the number of moles of H2SO4 and multiplied that by 2 to find number of moles of H+ and found how many mL of KOH gives this many moles. But what I don't understand is that aren't H+ and OH- already in a 1:1 ratio? Why do you have to multiply H+ by 2? Do you just disregard molar ratios for pH calcs?
Hopefully that made sense! Thank you
I thought I"d give it a go
First we work out the moles of H2SO4 .
Moles = concentration * volume = 0.0005*0.015= 0.0000075
Since this is a 1:2 ratio of H2SO4 to KOH
Multiply 0.0000075 by 2 = 0.000015 mols of KOH
Volume = moles/concentration = 0.000015/0.005= 0.003
which is 3 ml
I'm not sure if I did it correctly though....tell me if I'm wrong because I totally could be , and I'm also not sure how they calculated H+
I just worked out the moles of H2SO4 rather than the moles of H+, and then multiplied by 2 to find moles of KOH, since there's a 1:2 ratio.
And then from the moles of KOH I worked out the volume via volume = moles/concentration.
Maybe (idk tho) try to work with the moles of the substances rather than H+ and OH- in this case...? I have no clue, maybe you can post the answers for someone else to have a look?
whoops i was looking at 2nd last page rather than last, didn't notice someone already posted a reply