Yes, you are right but you need to tweek your thinking a little.
1. You have 4 significant figures in 25.00 and 3 in 1.00g; therefore, you are allowed only 3 s.f. in the answer. Thus, your answer of 0.7128 should be rounded to three s.f. of 0.713M. Some profs will count 0.7128 M wrong.
2. The reaction is KOH + HCl ==> KCl + H2O
Therefore, you properly converted 1.00 g KOH to mols KOH. The next step is to convert mols KOH to mols HCl. In this case the ratio in the reaction is 1:1 (1 mol KOH to 1 mol HCl) and it is obvious that mols KOH = mols HCl and the conversion is automatic. I'm just bringing this up for cases when the ratio in the titration is not 1:1. In
Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O
mols Ba(OH)2 = M x L
mols HCl = 2x mols Ba(OH)2
Then M HCl = mols HCl/L HCl
I'm not trying to confuse you and probably you already know all of this; I'm just trying to make sure you don't leave out that step when it is necessary.
Calculate the molarity of the HCl solution if 25.00ml of the acid are required to neutralize 1.00g of each of the following
1) KOH
2)Ca(OH)2
I calculated the molar mass of KOH which was 56.11
Then I took 1.00g/56.11 = 0.0178mol
To find out the molarity I did 0.0178/0.025=0.7128M
I just want to know what I'm doing is right.
1 answer
1 answer