If 50 ml of .50M KOH is added to 75 ml of .20M H2SO4, will solution be neutral, acidic, or basic?

Question

I know the balanced equation is 2KOH+H2SO4-> K2SO4+2H2O but i'm not quite sure how to solve it. Thanks for any input you give!

Answer 1

Your final solution would be acidic. Here's how to work this problem out: you have (50mL x 0.50 M) = 25 mmol KOH (and therefore OH-); then you have (75mL x 0.2M) = 15mmol H2SO4, (but 1mmol H2SO4 = 2mmol H+, so you have a total of 30mmol H+). 25 mmol OH will neutralize only 25mmol H+, therefore you have an excess of 5mmol H+, and the end solution is acidic.

Answer 2

the solution would be basic. the moles of KOH in solution is its molarity multiplied by the total volume in solution in liters. 50ml + 75ml = .125 liters...
so KOH = .50 M * .125 = .0625 moles
H2SO4 = .20 M * .125 = .025 moles
The H2SO4 would actually have 2 equilibrium points since it has 2 H's. But the second H always dissociates less than the first one and the amount of OH in excess would also neutralize them. there would still be OH in excess. it would result in a basic solution.

Answer 3

I would guess basic
KOH in this problem is more than twice as strong as the acid
The acid has more volume , but not 2 1/2 times the .2M to equal The .5 M KOH

Answer 4

this is uncomplicated. we've .25 moles oh koh and a million.5 moles of acid now acid will neutralize the backside as much as .25 moles of acid. yet acid is dibasic so this is going to neutralize 0.5 the fee. so ensuing mixture might desire to be acidic. reacn 2KOH + H2SO4 - K2SO4 +2H2O

Answer 5

mole of KOH=Volume(KOH)*C(KOH)=0.05*0.5=0.025 mol
mole of H2SO4=Volume(H2SO4)*C(H2SO4)=0.075*0.2=0.015 mol
mole of KOH/mole of H2SO4=0.025/0.015=1.67<2
-->H2SO4 is in excess
-->the solution would be acidic

Answer 6

Wouldn't OH make it a base?