50.0 ml of .30 KOH are required to titrate 60.0 ml of H2SO4. What is the M of the H2SO4?
Is .13 correct?
To two significant figures, yes. If that 0.30 was really 0.300 (and you just omitted a zero) then you could have 3 s.f. and that would be 0.125M. As posted, however, 0.13 is right.
Well, you know what they say, chemistry can be quite titrating! Let me calculate the molarity of the H2SO4 for you with a splash of humor.
To find the molarity, we can use the formula: Molarity = moles of solute / volume of solution in liters.
Given that 50.0 mL of 0.30 M KOH is required to titrate 60.0 mL of the mysterious H2SO4, we can set up a little equation:
(0.30 M KOH)(0.050 L KOH) = (x M H2SO4)(0.060 L H2SO4)
Let's do some math magic to solve this equation!
(0.30 M)(0.050 L) / 0.060 L = x M
By doing the calculations, we find that the molarity of H2SO4 is approximately 0.25 M. So, sorry to burst your bubble, but 0.13 M is not correct in this case. Keep those chem jokes coming!
To find the molarity of H2SO4, you can use the formula for the balanced chemical equation of the reaction between KOH and H2SO4, which is:
2 KOH + H2SO4 -> K2SO4 + 2 H2O
From the equation, you can see that the molar ratio between KOH and H2SO4 is 2:1. This means that for every 2 moles of KOH, you need 1 mole of H2SO4.
Given that 50.0 mL of 0.30 M KOH is required to titrate 60.0 mL of H2SO4, we can start by calculating the moles of KOH used:
moles of KOH = volume (in L) × molarity
moles of KOH = 50.0 mL × (1 L/1000 mL) × 0.30 mol/L
moles of KOH = 0.015 mol
Since the molar ratio between KOH and H2SO4 is 2:1, this means that the moles of H2SO4 used in the titration is half of the moles of KOH:
moles of H2SO4 = 0.015 mol / 2
moles of H2SO4 = 0.0075 mol
Now we can calculate the Molarity of H2SO4 by dividing the moles of H2SO4 by the volume in liters:
Molarity of H2SO4 = moles of H2SO4 / volume (in L)
Molarity of H2SO4 = 0.0075 mol / 0.060 L
Molarity of H2SO4 = 0.125 M (rounded to three significant figures)
Therefore, the molarity (M) of the H2SO4 solution is not 0.13 M; it is 0.125 M.
To find the molarity (M) of the H2SO4, we can use the equation:
M1V1 = M2V2
where M1 is the molarity of KOH, V1 is the volume of KOH used, M2 is the molarity of H2SO4, and V2 is the volume of H2SO4.
Given:
V1 = 50.0 mL
V2 = 60.0 mL
M1 = 0.30 M
Let's plug in the values:
(0.30 M)(50.0 mL) = M2(60.0 mL)
Now we can solve for M2:
(0.30 M)(50.0 mL) = M2(60.0 mL)
15.0 mmol = 60.0 M2
M2 = 15.0 mmol / 60.0 mL
M2 = 0.25 M
So the molarity (M) of the H2SO4 is 0.25 M, not 0.13 M.