A sample of potassium hydrogen oxalate, KHC204, weighing 0.717 g, was dissolved in water and titrated with 18.47 mL of an NaOH solution. Calculate the molarity of the NaOH solution.
KHC2O4 + NaOH ==> NaKC2O4 + H2O
mols KHC2O4 = grams/molar mass = ?
mols NaOH = mols KHC2O4(see the coefficients in the balanced equation).
Then M NaOH = mols NaOH/L NaOH and solve for M
To calculate the molarity of the NaOH solution, we need to use the balanced equation for the reaction between NaOH and KHC204:
2NaOH + KHC204 -> 2H2O + Na2C204 + KOH
From the equation, we can see that 2 moles of NaOH react with 1 mole of KHC204. Therefore, the number of moles of NaOH can be calculated using the equation:
moles of KHC204 = mass / molar mass
Given that the mass of KHC204 is 0.717 g and its molar mass is 128.13 g/mol, we can calculate the moles of KHC204:
moles of KHC204 = 0.717 g / 128.13 g/mol = 0.005597 moles
Since 2 moles of NaOH react with 1 mole of KHC204, the number of moles of NaOH is equal to half the number of moles of KHC204. Therefore:
moles of NaOH = 0.005597 moles / 2 = 0.0027985 moles
Now, to calculate the molarity of the NaOH solution, we can use the equation:
molarity (M) = moles of solute / volume of solution (L)
Given that the volume of the NaOH solution is 18.47 mL, we need to convert it to liters:
volume of NaOH solution = 18.47 mL = 0.01847 L
Now, we can calculate the molarity of the NaOH solution:
Molarity of NaOH = 0.0027985 moles / 0.01847 L = 0.1515 M
Therefore, the molarity of the NaOH solution is 0.1515 M.
Note: It is important to double-check the calculations and ensure that the equation used is balanced and accurate.
To calculate the molarity of the NaOH solution, we need to use the balanced equation to determine the stoichiometry between KHC204 and NaOH.
The balanced equation between KHC204 and NaOH is:
KHC204 + NaOH -> NaKC204 + H2O
From the equation, we can see that one mole of KHC204 reacts with one mole of NaOH. Therefore, the number of moles of NaOH can be determined from the amount of KHC204 used in the titration.
First, we need to convert the mass of KHC204 to moles. The molar mass of KHC204 can be calculated as follows:
K (39.10 g/mol) + H (1.01 g/mol) + C (12.01 g/mol) + 2 * O (16.00 g/mol) + 4 * O (16.00 g/mol) = 128.13 g/mol
Moles of KHC204 = mass of KHC204 / molar mass of KHC204
= 0.717 g / 128.13 g/mol
= 0.00559 mol
Since the balanced equation tells us that one mole of KHC204 reacts with one mole of NaOH, the number of moles of NaOH can be determined as follows:
Moles of NaOH = 0.00559 mol
Now, we need to calculate the molarity of the NaOH solution using the volume of NaOH used in the titration.
Molarity (M) = Moles of solute / Volume of solution (in liters)
Volume of solution = 18.47 mL = 0.01847 L
Molarity of NaOH = Moles of NaOH / Volume of solution
= 0.00559 mol / 0.01847 L
= 0.302 M
Therefore, the molarity of the NaOH solution is 0.302 M.