Question

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

Answer 1

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)?

Answer 2

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.
<b>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</b>

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
<b>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. </b>

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)?

<b> 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.</b>