it takes 37.50 mL of 0.152 M sodium chromate to titrate 25.00 mL of silver nitrate. what is the molarity of the silver nitrate solution
37.5ml 1L .152 mol NaCr 1 mol AgNO3
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10^3ml 1L 1 mol Na2cr
1 L 25ml
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1 mole AgNo3 10^3ml
37.5 x .152 / 25
To find the molarity of the silver nitrate solution, we can use the concept of stoichiometry. The balanced equation for the reaction between sodium chromate (Na2CrO4) and silver nitrate (AgNO3) is:
2 AgNO3 + Na2CrO4 → Ag2CrO4 + 2 NaNO3
From the balanced equation, we can see that 2 moles of silver nitrate react with 1 mole of sodium chromate. Therefore, the number of moles of silver nitrate can be calculated as follows:
moles of silver nitrate = (moles of sodium chromate) / 2
We have the volume and molarity of the sodium chromate solution:
Volume of sodium chromate solution = 37.50 mL
Molarity of sodium chromate solution = 0.152 M
First, we need to convert the volume of sodium chromate solution to the number of moles. We know that:
moles = volume (in L) x molarity
Therefore, the moles of sodium chromate can be calculated as follows:
moles of sodium chromate = (37.50 mL / 1000 mL/L) * (0.152 mol/L)
Now let's calculate the moles of silver nitrate:
moles of silver nitrate = (moles of sodium chromate) / 2
Finally, we find the molarity of the silver nitrate solution by dividing the moles of silver nitrate by the volume of silver nitrate solution:
Molarity of silver nitrate solution = (moles of silver nitrate) / (volume of silver nitrate solution in L)
Please provide the volume of silver nitrate solution, so we can proceed with the calculation.
To determine the molarity (M) of the silver nitrate (AgNO3) solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between sodium chromate (Na2CrO4) and silver nitrate (AgNO3).
The balanced chemical equation for the reaction is as follows:
2 AgNO3 + Na2CrO4 → Ag2CrO4 + 2 NaNO3
From the given information, we know that it takes 37.50 mL of 0.152 M sodium chromate to titrate 25.00 mL of silver nitrate. Since the stoichiometric ratio between sodium chromate and silver nitrate is 1:2 (from the balanced equation), we need to account for this ratio.
Step-by-step solution:
1. Convert the volume of silver nitrate solution from mL to liters:
V(AgNO3) = 25.00 mL = 0.02500 L
2. Determine the number of moles of sodium chromate used in the reaction:
n(Na2CrO4) = C(Na2CrO4) * V(Na2CrO4)
= 0.152 M * 0.03750 L
= 0.0057 moles
3. Use the stoichiometry of the balanced equation to find the number of moles of silver nitrate:
n(AgNO3) = 2 * n(Na2CrO4)
= 2 * 0.0057 moles
= 0.0114 moles
4. Calculate the molarity of the silver nitrate solution using the number of moles and volume:
M(AgNO3) = n(AgNO3) / V(AgNO3)
= 0.0114 moles / 0.02500 L
= 0.456 M
Therefore, the molarity of the silver nitrate solution is 0.456 M.