A 10.0 mL sample of 3.00 M KOH(aq) is transferred to a 250.0 ml
volumetric flask and diluted to the mark. It was found that 38.5 ml of
this diluted solution was needed to react the stoichimetric point in a
titration of 10.0 mL of a phosphoric acid, H3PO4, solution. The reaction
is:
3 KOH(aq) + H3PO4(aq) ® K3PO4(aq) + 3 H2O(l)
a) Calculate the molarity of H3PO4 in the original solution. [3]
b) Calculate the percent , by mass, of H3PO4 in the original solution.
Assume the density of the acid is 1.00 g/mL. [2]
A) .154 M
B) Not sure, 5.8 %
woops i mesed up
here's my answer
A .120
b 1.51%
question 3: Calculate the density of COCl2, a poisonous gas at 27.0oC and 733 Torr. [3]
i got 3.87 g/v
I have 3.8758 g/L which rounds to 3.88 g/L. Note g/L and g/v.
thanks
looks like im on track
To solve this problem, we need to use the principles of stoichiometry and titration. Let's break it down into steps:
Step 1: Determine the moles of KOH transferred to the volumetric flask.
To do this, we use the formula:
moles = concentration (M) × volume (L)
Given:
concentration of KOH = 3.00 M
volume of KOH transferred = 10.0 mL
First, convert the volume to liters:
10.0 mL = 10.0 mL × (1 L / 1000 mL) = 0.0100 L
Now, calculate the moles of KOH transferred to the flask:
moles of KOH = 3.00 M × 0.0100 L = 0.0300 mol
Step 2: Determine the moles of H3PO4 that reacted in the titration.
According to the balanced equation, the mole ratio between KOH and H3PO4 is 3:1. This means that for every 3 moles of KOH, there is 1 mole of H3PO4.
Given:
volume of diluted KOH used in the titration = 38.5 mL
Convert the volume to liters:
38.5 mL = 38.5 mL × (1 L / 1000 mL) = 0.0385 L
Now, use the mole ratio to calculate the moles of H3PO4:
moles of H3PO4 = (0.0300 mol KOH) × (1 mol H3PO4 / 3 mol KOH) = 0.0100 mol H3PO4
Step 3: Calculate the molarity of H3PO4 in the original solution.
To find the molarity, we need to consider the original volume of the H3PO4 solution used in the titration.
Given:
volume of H3PO4 used in the titration = 10.0 mL
Convert the volume to liters:
10.0 mL = 10.0 mL × (1 L / 1000 mL) = 0.0100 L
Now, use the moles of H3PO4 and the volume to calculate the molarity:
molarity of H3PO4 = (0.0100 mol H3PO4) / (0.0100 L) = 1.00 M
So, the molarity of H3PO4 in the original solution is 1.00 M.
For part (b), calculating the percent, by mass, of H3PO4 in the original solution involves finding the mass of H3PO4 in the 10.0 mL of solution used in the titration.
Given:
density of the acid = 1.00 g/mL
volume of H3PO4 used in the titration = 10.0 mL
The mass of the H3PO4 used in the titration is equal to the volume multiplied by the density:
mass of H3PO4 = 10.0 mL × 1.00 g/mL = 10.0 g
To calculate the percent, by mass, divide the mass of H3PO4 by the total mass of the 10.0 mL solution and multiply by 100:
percent, by mass, of H3PO4 = (10.0 g / 10.0 g) × 100% = 100%
However, since the question specifically asks for the percent, by mass, of H3PO4 in the original solution, we need to account for dilution.
The original solution was diluted 25 times when transferred to the 250.0 mL volumetric flask. So, the mass of H3PO4 in the original solution is:
mass of H3PO4 in original solution = (10.0 g / 25) = 0.400 g
Now, calculate the percent, by mass, using the mass of H3PO4 in the original solution and the mass of the 10.0 mL solution:
percent, by mass, of H3PO4 = (0.400 g / 10.0 g) × 100% = 4.00%
Therefore, the correct answers are:
a) The molarity of H3PO4 in the original solution is 1.00 M.
b) The percent, by mass, of H3PO4 in the original solution is 4.00%.