50 ml of 0.10 M NaOH completely neutralizes .30 ml of hydrochloric acid. Calculate the concentration of HCL.
-
.1 M NaOH = .1mol/L NaOH
Using 50 mL (.05L), .05L NaOH * .1 mol/L NaOH = .005 mol NaOH
If you're completely neutralizing this, that means that mols NaOH = mols HCl, therefore .005 mol NaOH = .005 mol HCl
And since you know the starting volume for HCl (.30 mL? or do you mean 30 mL?) all you have to do is convert the mL to L and divide the # of mols (.005) by that value
Using 50 mL (.05L), .05L NaOH * .1 mol/L NaOH = .005 mol NaOH
If you're completely neutralizing this, that means that mols NaOH = mols HCl, therefore .005 mol NaOH = .005 mol HCl
And since you know the starting volume for HCl (.30 mL? or do you mean 30 mL?) all you have to do is convert the mL to L and divide the # of mols (.005) by that value
-
Yep!
Report Abuse
-
I assume 30 ml HCl, instead of .30ml:
Molarity of HCl: (.05 * .1)/ .03 = .17 mol
Molarity of HCl: (.05 * .1)/ .03 = .17 mol