hey quick help plz

[mdgurl07]

i have this problem and idk what to do?

can someone plz help me....

 

--> A 36.0 mL sample of aqueous sulfuric acid was titrated with 0.250 molar KOH(aq) until the chemical indicator signaled that the solution was exactly neutralized. The mixture was then carefully evaporated to dryness, then the residue was dried in a drying oven. The residue was then weighed and a value of 861 mg was obtained for its mass. Calculate a value for the quantity of the potassium hydroxide solution that was used in the titration from this data.

[insane_alien]

okay, work out how many moles of sulphuric acid you had from the amount of potassium sulphate you had left over then work out how much of the solution you needed.

 

if you can show us what you've done already that would also help.

[mdgurl07]

oh k well lets see if i can do that much thanks a bunch!

[mdgurl07]

k lets see i did this:

H2SO4 + 2 KOH -> K2SO4 + 2 H20

 

(0.861g of KOH)(1 MOL koh/112.2114g KOH)(1 mol H2SO4/2 mol KOH)=3.8 x 10^-3 mol H2SO4

[mdgurl07]

now wat do i do is that even correct?

[insane_alien]

the residue isn't KOH its K2SO4. you'll need to that calculation again. i think the method is right but your notation is a bit unreadable to me.

 

okay, after that you can work out how many moles of KOH. you know the molarity of the solution. from these you should be able to work out the answer.