A 20.00 ml stock sample of HBr is diluted to 50 mL. If 18.76 ml of 0.1345 M KOH was required to reach the equivalence point, what is the concentration of the stock HBr solution?
Is my answer correct?
Titration problems can be solved with the MV=MV equation. The Molarity * Volume of one solution = the Molarity * Volume of the other solution.
In this case, I think the volume it's diluted to is throw away info meant to confuse you. You neutralized 20mL of stock solution with some water added to it. The water doesn't matter. So throw out that 50mL number.
So (???M)*(20mL) = (0.1345M)*(18.76mL)
The HBr is 0.1262M
The digits are correct but I need to say a couple of things. First, The MV = MV is true ONLY (AND ONLY) when the reaction is 1 mol to 1 mol ratio (2:2 is ok since that is still 1:1). So don't forget. Second, I don't know how many significant figures you had in the original problem but ther answer of 0.1262 probably is too many.
How about this answer?
HBr + KOH > H2O + KBr
We know the amount of potassium hydroxide in moles by simply calculating it based on the information given, with the formula:
c= n/v ---> n = c*v
After figuring out the number of moles of potassium hydroxide, we can very easily determine the number of moles of hydrobromic acid required for neutralization- it's equal, as seen in the balanced equation.
After that, we need to know the concentration of the acid in a 20.00 mL solution. Again, just apply the formula c= n/v. Then you have the stock concentration.
Don't let the "diluted to 50 mL" part throw you off. All that it does is add more solution, the chemical reaction will still happen (slower I suppose). There must be equal moles of both chemicals.
All of that is ok. And you are right about the dilution. The chemicals react with mols of one neutralizing mols of the other and the amount of water doesn't make any difference. My students always said, "But you've dilute the first chemical so it's not as concentrated." I always replied, "But it diluted the second chemical when it was added by the same amount."
I do the ones that are not 1:1 this way.
Determine the molarity of KOH if 20 mL of KOH reacted exactly with 50 mL of 0.2 M H3PO4.
3KOH + H3PO4 ==> K3PO4 + 3H2O
mols H3PO4 = M x L = 0.2 x 0.050 = 0.010
convert mols H3PO4 to mols KOH.
0.010 mol H3PO4 x (3 mol KOH/1 mol H3PO4) = 0.010 x 3/1 = 0.030 mol KOH.
convert 0.030 mol KOH to M from
L x M = mols
0.020 x M = 0.030
M KOH = 0.030/0.020 = 1.5 M KOH.
hey so is the answer correct for the question?
Your approach is correct, but there is a small error in the final calculation. Here's the correct calculation:
Using the equation MV = MV, we can set up the equation:
(Molarity of HBr) * (Volume of HBr) = (Molarity of KOH) * (Volume of KOH)
Let's plug in the values we have:
(Molarity of HBr) * (20.00 mL) = (0.1345 M) * (18.76 mL)
Now let's solve for the concentration of HBr:
(Molarity of HBr) = (0.1345 M) * (18.76 mL) / (20.00 mL)
Molarity of HBr ≈ 0.1258 M
So, the concentration of the stock HBr solution is approximately 0.1258 M.
Your answer of 0.1262 M is slightly off due to a rounding error.