Posted by Dannie on .
A solution is prepared by adding 50 mL of .050 M HCL to 150 mL of .10 M HNO3. Calculate the concentrations. I know I need to find the [H+] and [OH-] but im not sure how.
moles HNO3 = M x L = ??
M HNO3 = moles/total L.
moles HCl = M x L = xx
M HCl = moles/total L.
Another way, to me less complicated; however, it gets away from the defnitions of M = moles/L which is all the above. Just the definition. The other way is a dilution method.
(HNO3) = 0.1M x (150 mL/200 mL) = ?M
(HCl) = 0.05M x (50 mL/200 mL) = ?M
Try it both ways. You should get the same answer either way.
The balanced chemical equation for reaction of Ba(OH) and HCl is as follows:
Ba(OH) + 2HCl BaCl + 2H O
1 mole Ba(OH) produce 2 moles OH ions.
Number of moles of OH ion in solution = concentration of Ba(OH) x volume of Ba(OH)
= 0.1 mol L x 0.03 L x 2
= 0.006 mol
Number of moles of H ion in solution = Concentration of HCl x volume of HCl
= 0.05 mol L x 0.02 L
= 0.001 mol
Now after mixing 0.001 mol H will reacts with 0.001 mol OH .
Number of moles of OH left unreacted = 0.006 - 0.001
= 0.005 mol
Total volume of solution = 20 ml + 30 ml = 50 ml = 0.05 L
Concentration of OH ion in ánal solution = 0.005 mol / 0.05 L = 0.1 mol L .