Question

What volume of a 1.5 M KOH solution is requred to titrate 30 mL of a 4.5 M H2SO4?

H2SO4+2KOH --> K2SO4+2H2O

Answer 1

You will need twice as many moles of KOH as H2 SO4 for the neutralization reaction to be in molar proportions.

Answer 2

so what would the volume be? the answers are...
a. 158 mL
b. 125mL
c. 200mL
d.45.0 mL
e. None of the above

Answer 3

Do you know how to find the moles of H2SO4 you have? (I wont just give you the answer. It is against policy)

Answer 4

isnt the moles oh H2SO4 given in the question? 4.5?
@JoannaBlackwelder

Answer 5

No that is the molarity (M). Molarity is in moles/Liter

Answer 6

Im not really sure how to start it off @JoannaBlackwelder

Answer 7

Ok. Since you have 30 mL of 4.5 M H2SO4 (that is 4.5 mol H2SO4/ Liter), convert 30 mL to L.

Answer 8

Then take the number of liters you have and convert it to moles of H2SO4 by multiplying by the molarity. This cancels out liters and leaves you with moles.

Answer 9

Does that make sense so far?

Answer 10

so 4.5=.03/x which is .0067

Answer 11

No, 0.03 L *(4.5 mol/1 L) = 0.135 mol

Answer 12

It is like converting units.

Answer 13

ok i understand so whats next

Answer 14

We can see from the reaction that you need 2 moles of KOH for every 1 mol H2SO4 for the reaction to be at stoichiometric ratios.

Answer 15

So, 0.135 mol H2SO4 *(2 mol KOH/1 mol H2SO4) =

Answer 16

.27

Answer 17

0.27 moles of KOH needed to completely neutralize.

Answer 18

Right. Do you understand that a titration is a neutralization reaction?

Answer 19

yes

Answer 20

Great! So, now can you calculate the volume of KOH (for 1.5 M KOH)?

Answer 21

.27/1.5?

Answer 22

Yup :)

Answer 23

so would the answer be none of the above
a. 158 mL
b. 125mL
c. 200mL
d.45.0 mL
e. None of the above

Answer 24

because it would be 180mL

Answer 25

Yes.

Answer 26

Correct.

Answer 27

thank you so much

Answer 28

No worries