Posted by **Paris** on Saturday, June 25, 2011 at 9:27pm.

A volume of 60.0 of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 25.2 mL of 1.50M H2SO4 was needed? The equation is 2KOH+H2SO4-->K2SO4+2H2O. I cant seem to get the right answer so please, if you know how to do this explain it fully and work out the problem completely with numbers and equations.

- Chemistry -
**DrBob222**, Saturday, June 25, 2011 at 10:18pm
Has it occurred to you that if you had posted your work I could have found the error in half the time it takes to work the problem.

H2SO4 + 2KOH ==> K2SO4 + 2H2O

How many moles H2SO4 were used? That is M x L = moles H2SO4.

Now convert that to moles KOH. From the equation, it takes 2 moles KOH to equal 1 mole H2SO4; therefore, moles H2SO4 x 1/2 = moles KOH.

Now M KOH = moles KOH/L KOH.

## Answer this Question

## Related Questions

- Chemistry - A volume of 40.0 of aqueous potassium hydroxide (KOH) was titrated ...
- Chemistry - A volume of 70.0mL of aqueous potassium hydroxide(KOH) was titrated ...
- Chemistry - A volume of 80.0mL of aqueous potassium hydroxide (KOH) was titrated...
- pcc - A volume of 40.0mL of aqueous potassium hydroxide (KOH) was titrated ...
- chemistry 112 - A volume of 60.0mL of aqueous potassium hydroxide (KOH) was ...
- Chemistry - A volume of 50.0 mL of aqueous potassium hydroxide was titrated ...
- chemistry - A volume of 40.0 of aqueous potassium hydroxide was titrated against...
- Chemistry - A volume of 60.0mL of aqueous potassium hydroxide (KOH ) was ...
- chemistry - 35.00mL of 0.250M KOH solution is required to completely neutralize ...
- chemistry - What are the products formed when sulfuric acid solution reacts with...

More Related Questions