1-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.
2-A 35 mL drinking water sample, whose pH is buffered to 10, is titrated with 18.41 mL 4.252x10-3 M EDTA beside the eriochrome T indicator, express the hardness of the water
1-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.
KHC2O4 + NaOH ==> NaKC2O4 + H2O
I assume the KHC2O4 is being used as a primary standard. Then mols KHC2O4 = grams/molar mass = 0.717/molar mass = ?
From the equation, mols NaOH = mols KHC2O4
M NaOH = mols/L = mols from above/0.01847
2-A 35 mL drinking water sample, whose pH is buffered to 10, is titrated with 18.41 mL 4.252x10-3 M EDTA beside the eriochrome T indicator, express the hardness of the water
mols EDTA = M x L = ?
1 mol EDTA = 1 mol Mg and Ca ions (hardness); therefore,
mols EDTA = mols CaCO3
ppm hardness = mols CaCO3 x 100.09 g/mol x 1000 g/mg x (1/0.035 L) = ?
Post your work if you get stuck.
The standard solution of 0.0500 M K + is required for the calibration solution in the determination of potassium by flame photometric method. How to prepare 50 mL of this solution from primary standard K2CO3 (138,205 g / mol)?
You want 0.0500 M K^+ which is equivalent to 0.02500 M K2CO3.since K2CO3 conains 2 mols K for each mols K2CO3.
mols K2CO3 needed = M x L = 0.025 x 0.05 = ?
grams K2CO3 = mols K2CO3 x molar mass K2CO3 = ?
Post your work if you need more help.
The standard solution of 0.0500 M K + is required for the calibration solution in the determination of potassium by flame photometric method. How to prepare 50 mL of this solution from primary standard K2CO3 (138,205 g / mol)?
1) To calculate the molarity of the NaOH solution, we can use the concept of stoichiometry and the equation for the reaction between KHC2O4 and NaOH:
KHC2O4 + NaOH -> NaKC2O4 + H2O
First, we need to determine the number of moles of KHC2O4 used in the titration. We can do this by using the given mass and the molar mass of KHC2O4:
Mass of KHC2O4 = 0.717 g
Molar mass of KHC2O4 = 39.1 g/mol (K) + 1.01 g/mol (H) + 2 * 12.01 g/mol (C) + 4 * 16 g/mol (O) = 128.13 g/mol
Number of moles of KHC2O4 = Mass / Molar mass = 0.717 g / 128.13 g/mol = 0.0056 mol
From the balanced equation, we know that the stoichiometric ratio between KHC2O4 and NaOH is 1:1. This means that the number of moles of NaOH used in the titration is also 0.0056 mol.
Now, we can calculate the molarity of the NaOH solution using the equation:
Molarity (M) = Number of moles / Volume (L)
Volume of NaOH solution = 18.47 mL = 18.47 / 1000 L = 0.01847 L
Molarity of NaOH solution = 0.0056 mol / 0.01847 L = 0.303 M
Therefore, the molarity of the NaOH solution is 0.303 M.
2) To express the hardness of the water, we need to determine the concentration of calcium and magnesium ions present.
The reaction between the calcium and magnesium ions with EDTA can be written as:
Ca2+ (aq) + EDTA4- (aq) -> Ca(EDTA)2- (aq)
Mg2+ (aq) + EDTA4- (aq) -> Mg(EDTA)2- (aq)
Given that the pH is buffered to 10, the eriochrome T indicator is used to determine the endpoint of the titration. At pH 10, the indicator changes color from blue to red. This indicates the formation of the calcium-EDTA complex and the endpoint of the titration.
Since the concentration of EDTA is given as 4.252x10-3 M and the volume used in the titration is 18.41 mL, we can calculate the number of moles of EDTA used:
Number of moles of EDTA = Molarity x Volume
= 4.252x10-3 M x 0.01841 L
= 7.8302x10-5 mol
From the balanced equation, we know that the stoichiometric ratio between Ca2+ and EDTA is 1:1. Therefore, the number of moles of calcium ions present in the water sample is also 7.8302x10-5 mol.
The hardness of water is usually expressed in terms of calcium carbonate (CaCO3) equivalents. Since the molar mass of calcium ion (Ca2+) is 40.08 g/mol and the molar mass of calcium carbonate (CaCO3) is 100.09 g/mol, we can calculate the hardness of the water sample in terms of CaCO3 equivalents:
Hardness (CaCO3 equivalents) = Number of moles x Molar mass (CaCO3) / Molar mass (Ca2+)
= 7.8302x10-5 mol x 100.09 g/mol / 40.08 g/mol
= 1.952 g/L
Therefore, the hardness of the water sample is 1.952 g/L in terms of CaCO3 equivalents.
To solve both questions, we need to use the concept of stoichiometry, which relates the relative quantities of substances involved in a chemical reaction.
1. Calculation of the molarity of NaOH solution:
Let's start by writing the balanced chemical equation for the reaction between NaOH and KHC2O4:
NaOH + KHC2O4 -> NaKC2O4 + H2O
From the balanced equation, we can see that the mole ratio between NaOH and KHC2O4 is 1:1, meaning that one mole of NaOH reacts with one mole of KHC2O4.
Given that the mass of KHC2O4 is 0.717 g, we first need to convert this mass to moles using the molar mass of KHC2O4. The molar mass of KHC2O4 is:
K: atomic mass = 39.1 g/mol
H: atomic mass = 1.01 g/mol
C: atomic mass = 12.01 g/mol
O: atomic mass = 16.00 g/mol
Total molar mass = (39.1 + 1.01 + (12.01 * 2) + (16.00 * 4)) g/mol = 128.12 g/mol
Number of moles of KHC2O4 = mass / molar mass = 0.717 g / 128.12 g/mol
Now, we can use the volume and molarity of NaOH solution to calculate its molarity:
Molarity of NaOH = (number of moles of KHC2O4) / (volume of NaOH in liters)
Note: To use the volume in liters, we need to convert the given volume of 18.47 mL to L:
Volume of NaOH in L = 18.47 mL * (1 L / 1000 mL)
Finally, substitute the values into the equation:
Molarity of NaOH = (0.717 g / 128.12 g/mol) / (18.47 mL * (1 L / 1000 mL))
2. Calculation of water hardness:
The hardness of water can be determined by titration using the chelating agent EDTA (ethylene diamine tetraacetic acid). In this case, we are given that the pH of the water sample is buffered to 10, and it is titrated with 18.41 mL of 4.252x10-3 M EDTA using the eriochrome T indicator.
The hardness of the water can be expressed in terms of calcium carbonate (CaCO3) equivalents. The balanced equation for the reaction between EDTA and Ca2+ is:
Ca2+ + EDTA -> Ca-EDTA
From the balanced equation, we can see that the mole ratio between Ca2+ and EDTA is 1:1. Hence, the number of moles of Ca2+ can be determined from the moles of EDTA used.
To calculate the number of moles of Ca2+, we first need to calculate the moles of EDTA used using the given volume and molarity:
Number of moles of EDTA = volume of EDTA in L * molarity of EDTA
Now, we can use the stoichiometry to relate the moles of EDTA used to the moles of Ca2+:
Number of moles of Ca2+ = number of moles of EDTA
Finally, we can express the hardness of the water sample in terms of CaCO3 equivalents using the molar mass of CaCO3:
Hardness of water in CaCO3 equivalents (mg/L) = (number of moles of Ca2+) * (molar mass of CaCO3) * 1000 / volume of water sample in L
Note: The molar mass of CaCO3 is:
Ca: atomic mass = 40.08 g/mol
C: atomic mass = 12.01 g/mol
O: atomic mass = 16.00 g/mol
Total molar mass = (40.08 + (12.01 * 1) + (16.00 * 3)) g/mol = 100.09 g/mol