A sample of potassium hydrogen oxalate, KHC2O4, 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.
First, determine the molarity of KHC2O4:
0.717g*(1 mol/128.12g)=moles of KHC2O4
18.47 mL=0.01847L
Molarity of KHC2O4=moles of KHC2O4/0.01847L
The reaction is a 1 mole to 1 mole reaction. Therefore, moles of KHC2O4=moles of NaOH.
So, Molarity of KHC2O4=Molarity of NaOH
***Answer contains three significant figures.
To calculate the molarity of the NaOH solution, we need to use the balanced chemical equation for the reaction between potassium hydrogen oxalate (KHC2O4) and sodium hydroxide (NaOH).
The balanced equation is:
2KHC2O4 + 2NaOH -> 2H2O + Na2C2O4 + 2KOH
From the balanced equation, we can see that the mole ratio between KHC2O4 and NaOH is 2:2, or 1:1.
First, let's calculate the number of moles of KHC2O4:
Molar mass of KHC2O4 = (1 x 39.10 g/mol) + (1 x 1.01 g/mol) + (2 x 12.01 g/mol) + (4 x 16.00 g/mol)
= 128.13 g/mol
Number of moles of KHC2O4 = Mass / Molar mass
= 0.717 g / 128.13 g/mol
= 0.00559 mol
Since the mole ratio between KHC2O4 and NaOH is 1:1, the number of moles of NaOH is also 0.00559 mol.
Next, let's calculate the molarity of the NaOH solution:
Molarity (M) = Number of moles / Volume (L)
Volume is given as 18.47 mL, but we need to convert it to liters:
Volume (L) = 18.47 mL / 1000 mL/L
= 0.01847 L
Molarity (M) = 0.00559 mol / 0.01847 L
= 0.302 M
Therefore, the molarity of the NaOH solution is 0.302 M.
To calculate the molarity of the NaOH solution, we need to use the balanced chemical equation and the stoichiometry of the reaction between potassium hydrogen oxalate (KHC2O4) and sodium hydroxide (NaOH).
The balanced equation for the reaction is:
2KHC2O4 + 2NaOH -> 2H2O + 2CO2 + Na2C2O4
From the balanced equation, we can determine the stoichiometry of the reaction. It tells us that 2 moles of KHC2O4 react with 2 moles of NaOH to produce 2 moles of water (H2O), 2 moles of carbon dioxide (CO2), and 1 mole of sodium oxalate (Na2C2O4).
Now, let's calculate the number of moles of KHC2O4 in the sample:
Mass of KHC2O4 = 0.717 g
Molar mass of KHC2O4 = 39.10 g/mol + 1.01 g/mol + (12.01 g/mol x 2) + (16.00 g/mol x 4) = 128.09 g/mol
Moles of KHC2O4 = Mass of KHC2O4 / Molar mass of KHC2O4
= 0.717 g / 128.09 g/mol
= 0.00559 mol
Since the stoichiometry of the balanced equation is 1:1 between KHC2O4 and NaOH, the number of moles of NaOH used in the titration is also 0.00559 mol.
Now, we can calculate the molarity of the NaOH solution using the formula:
Molarity (M) = Moles of solute / Volume of solution (in liters)
Volume of solution = 18.47 mL = 0.01847 L
Molarity of NaOH solution = 0.00559 mol / 0.01847 L
= 0.302 M
Therefore, the molarity of the NaOH solution is 0.302 M.