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.
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.
Yes, your calculations for the molarity of the HCl solution look correct. Let's go through it step by step to verify:
1) For the reaction between HCl and KOH, the balanced equation is:
HCl + KOH → KCl + H2O
Given that 25.00 ml of HCl is required to neutralize 1.00 g of KOH, you correctly calculated the number of moles of KOH:
1.00 g / 56.11 g/mol = 0.0178 mol
Next, you divided the number of moles of KOH by the volume of HCl used (in liters) to find the molarity:
0.0178 mol / 0.025 L = 0.7128 M
So, the molarity of the HCl solution required to neutralize 1.00 g of KOH is indeed 0.7128 M.
For the second part of your question:
2) For the reaction between HCl and Ca(OH)2, the balanced equation is:
2HCl + Ca(OH)2 → CaCl2 + 2H2O
Make sure you take into account the stoichiometry of the reaction. Since 2 moles of HCl are required to neutralize 1 mole of Ca(OH)2, you need to adjust the calculation slightly.
The molar mass of Ca(OH)2 is 74.09 g/mol. So, 1.00 g of Ca(OH)2 is equal to:
1.00 g / 74.09 g/mol = 0.0135 mol
Since 2 moles of HCl are required to neutralize 1 mole of Ca(OH)2, you would then need double the amount of moles of HCl:
0.0135 mol * 2 = 0.027 mol
Finally, divide the number of moles of HCl by the volume of HCl used (in liters) to find the molarity:
0.027 mol / 0.025 L = 1.08 M
Therefore, the molarity of the HCl solution required to neutralize 1.00 g of Ca(OH)2 is 1.08 M.