The pH of saturated Sr(OH)2 is found to be 13.12. A 15.0ml sample of saturated Sr(OH)2 is diluted to 250.0ml in a volumetric flask. A 15.0ml sample of the diluted Sr(OH)2 is transferred to a beaker, and some water is added. The resulting solution requires 29.3ml of a HCl solution for its titration.
What is the molarity of this HCL solution?
The pH of the satd soln is 13.12; therefore, the pOH = 14 - 13.12 = 0.88
From this and pOH = -log(OH^-) you can find (OH^-) = about 0.12 (you need to do it exactly).
This was diluted from 15 mL to 250 so the (OH^-) in the 250 mL flask is 0.12 x 15/250 = about 0.0072 M (again you need to go through it to get a more exact answer).
So now you have a 15.0 mL sample of 0.0072 M OH^- titrated with 29.3 mL of ?M HCl.
mLbase x M base = mLacid x M acid.
Solve for M acid.
Well, looks like someone is diluting and titrating like a chemist!
To find the molarity of the HCl solution, we can use the equation:
M1V1 = M2V2
Where:
M1 = molarity of the Sr(OH)2 solution
V1 = volume of the Sr(OH)2 solution used
M2 = molarity of the HCl solution
V2 = volume of the HCl solution used
We know that the molarity of the Sr(OH)2 solution is saturated, which means it's fully dressed, so we don't need to worry about it here.
We also know that V1 (the volume of the Sr(OH)2 solution used) is 15.0 ml, and V2 (the volume of the HCl solution used) is 29.3 ml.
Now, we just need to rearrange the equation to solve for M2 (the molarity of the HCl solution):
M2 = (M1 * V1) / V2
But here's the thing, we don't have the molarity of the Sr(OH)2 solution. Tricky, right? It's like trying to find the height of a clown without knowing the size of their red nose!
So unfortunately, I can't give you an answer without that missing information. But don't worry, chemistry has a funny way of revealing its secrets when you least expect it!
To find the molarity of the HCl solution, we need to use the following equation:
M1V1 = M2V2
where:
M1 = Molarity of the HCl solution
V1 = Volume of the HCl solution used for titration (29.3 mL)
M2 = Molarity of the diluted Sr(OH)2 solution
V2 = Volume of the diluted Sr(OH)2 solution used for titration (15.0 mL)
We can rearrange the equation to solve for M1:
M1 = (M2 * V2) / V1
Now, let's calculate the molarity of the HCl solution.
The molarity of Sr(OH)2 can be calculated from its pH using the following equation:
pH = -log10([OH-])
Since Sr(OH)2 is a strong base, it will dissociate fully in water. Therefore, the concentration of hydroxide ions is equal to the concentration of Sr(OH)2:
[OH-] = [Sr(OH)2]
Using the fact that pOH + pH = 14, we can find the pOH:
pOH = 14 - pH
= 14 - 13.12
= 0.88
Using the equation [OH-] = 10^(-pOH), we can find the concentration of hydroxide ions:
[OH-] = 10^(-0.88)
= 0.1259 M
Since Sr(OH)2 is diluted, we need to calculate the concentration of Sr(OH)2 in the diluted solution:
M2 = [OH-] = 0.1259 M
Now, let's plug these values into the equation to find the molarity of the HCl solution:
M1 = (M2 * V2) / V1
= (0.1259 M * 15.0 mL) / 29.3 mL
= 0.0645 M
Therefore, the molarity of the HCl solution is 0.0645 M.
To determine the molarity of the HCl solution, we can use the equation:
M1V1 = M2V2
Where:
M1 = molarity of the HCl solution
V1 = volume of HCl solution used in the titration (in liters)
M2 = molarity of the Sr(OH)2 solution
V2 = volume of the Sr(OH)2 solution used in the titration (in liters)
From the given information:
V1 = 29.3 mL = 29.3/1000 L = 0.0293 L (converted to liters)
V2 = 15.0 mL = 15.0/1000 L = 0.0150 L (converted to liters)
Now, we need to calculate the molarity of the Sr(OH)2 solution.
First, let's find the moles of Sr(OH)2 in the 15.0 mL sample.
Molarity (M) = Moles (n) / Volume (V)
Since the Sr(OH)2 is saturated, we can assume it is fully dissociated.
Sr(OH)2 dissociates into 1 Sr2+ ion and 2 OH- ions.
So, the moles of Sr(OH)2 = moles of Sr2+ = 2 * moles of OH-
Given that the pH of saturated Sr(OH)2 is 13.12, we can find the pOH value:
pOH = 14 - pH
pOH = 14 - 13.12
pOH = 0.88
Now, convert pOH to OH- concentration using the equation:
OH- concentration = 10^(-pOH)
OH- concentration = 10^(-0.88) = 0.1258 M
So, the moles of OH- in the 15.0 mL sample = OH- concentration * volume of OH- solution (in L)
moles of OH- = 0.1258 * 0.0150 L = 0.00189 moles
Since Sr(OH)2 dissociates into 2 OH- ions, the moles of Sr(OH)2 = 0.00189 / 2 = 0.000943 moles
Now that we have the moles of Sr(OH)2 in the 15.0 mL sample, we can calculate the molarity of Sr(OH)2:
Molarity (M2) = Moles (n) / Volume (V)
Molarity (M2) = 0.000943 moles / 0.0150 L = 0.0629 M
Now, we can use the equation M1V1 = M2V2 to find the molarity of the HCl solution.
M1 * 0.0293 L = 0.0629 M * 0.0150 L
M1 = (0.0629 M * 0.0150 L) / 0.0293 L
M1 = 0.03236 M
Therefore, the molarity of the HCl solution is approximately 0.03236 M.