You perform the following titration: You placed 35 mL of H2S in a flask and placed it under a buret. The buret contains 0.243M Sr(OH)2 and you are going to titrate the H2S with the base. You titrate until the color changes. At that point, you record that you used 10.15mL of base to titrate. Calculate the unknown molarity of the H2S.
To calculate the unknown molarity of H2S, we can use the concept of stoichiometry. The balanced equation for the reaction between H2S and Sr(OH)2 is:
H2S + Sr(OH)2 -> SrS + 2H2O
From the equation, we can see that the mole ratio between H2S and Sr(OH)2 is 1:1. This means that for every 1 mole of H2S used, 1 mole of Sr(OH)2 will react.
First, let's calculate the number of moles of Sr(OH)2 used in the titration. We know the concentration and volume used.
Molarity of Sr(OH)2 = 0.243 M
Volume of Sr(OH)2 used = 10.15 mL = 0.01015 L (converted from mL to L)
Now, let's calculate the number of moles of Sr(OH)2 used:
Moles of Sr(OH)2 = Molarity * Volume
= 0.243 M * 0.01015 L
Next, since the mole ratio between H2S and Sr(OH)2 is 1:1, the number of moles of H2S used in the reaction will also be equal to the moles of Sr(OH)2 used.
Therefore, the number of moles of H2S used = 0.243 M * 0.01015 L
Now that we know the moles of H2S used and the volume of H2S used (35 mL = 0.035 L), we can calculate the molarity of H2S.
Molarity of H2S = Moles of H2S / Volume of H2S
= (0.243 M * 0.01015 L) / 0.035 L
Now, let's plug in the values to get the final answer:
Molarity of H2S = (0.243 * 0.01015) / 0.035 = 0.070257 M
So, the unknown molarity of H2S is approximately 0.070257 M.
Did you make up this problem or have it handed to you? However it was done you should recognize that H2S is a gas and isn't about to stand around in a titration vessel and wait to be titrated. Anyway, in the spirit of the problem here is how it is solved.
H2S + Sr(OH)2 ==> 2H2O + SrS
mols Sr(OH)2 = M x L = ?
Look at the equation and see that mols Sr(OH)2 = mols H2S
Then M H2S = mols H2S/L H2S.