What is the equivalent weight of: (a) Potassium tetraoxalate, KHC2O4•H2C2O4•2H2O, if the complete neutralization is assumed. (b) Sodium carbonate if it is titrated to the phenolphthalein end point with HCl. The reaction is Na2CO3 + HCl = NaCl + NaHCO3. (c) Sodium tetraborate decahydrate if the titration reaction is Na2 B4O7•10H2O + 2H+ + H2O = 4HBO2 + 2Na+? (d) Na3PO4•12H2O if it is titrated with phenolphthalein as an indicator to form HPO42-?
a. molar mass KHC2O4•H2C2O4•2H2O/3 = ?
b. molar mass Na2CO3/1 = ?
c. molar mass Na2 B4O7•10H2O /2 = ?
d. molar mass Na3PO4•12H2O/2 = ?
(a) To find the equivalent weight of potassium tetraoxalate, we need to determine the number of equivalents present in the formula. The formula for potassium tetraoxalate is KHC2O4•H2C2O4•2H2O.
The compound contains potassium (K), hydrogen (H), carbon (C), and oxygen (O). The atomic weights of these elements are as follows:
- Potassium (K): 39.1 g/mol
- Hydrogen (H): 1.0 g/mol
- Carbon (C): 12.0 g/mol
- Oxygen (O): 16.0 g/mol
To calculate the number of equivalents, we need to consider the species that can undergo neutralization. In this case, we have KHC2O4 and H2C2O4.
The equivalent weight of a compound is defined as the molecular weight divided by the number of equivalents.
For KHC2O4, we have one equivalent of K, one equivalent of H, and one equivalent of C2O4. Thus, the equivalent weight of KHC2O4 is:
(39.1 g + 1.0 g + 2*(12.0 g + 16.0 g)) / 1 equivalent = 39.1 g + 1.0 g + 56.0 g = 96.1 g/mol
For H2C2O4, we have two equivalents of H and two equivalents of C2O4. Thus, the equivalent weight of H2C2O4 is:
(2*(1.0 g) + 2*(12.0 g + 16.0 g)) / 1 equivalent = 2.0 g + 56.0 g = 58.0 g/mol
Now, let's calculate the equivalent weight of the compound, including the water molecules:
Equivalent weight of KHC2O4•H2C2O4•2H2O = (96.1 g + 58.0 g + 2*(18.0 g)) / 1 equivalent = 96.1 g + 58.0 g + 36.0 g = 190.1 g/mol
Therefore, the equivalent weight of potassium tetraoxalate is 190.1 g/mol.
(b) The balanced equation for the reaction between sodium carbonate (Na2CO3) and hydrochloric acid (HCl) is: Na2CO3 + 2HCl → 2NaCl + H2O + CO2
To find the equivalent weight of sodium carbonate, we need to consider the species that can undergo neutralization. In this case, we have Na2CO3.
The equivalent weight of a compound is defined as the molecular weight divided by the number of equivalents.
The formula weight of Na2CO3 can be calculated as follows:
(2*(22.99 g) + 12.01 g + 3*(16.00 g)) = 105.99 g/mol
The balanced equation shows that 2 moles of HCl react with 1 mole of Na2CO3, which means there is 1 equivalent of Na2CO3 per mole.
Therefore, the equivalent weight of sodium carbonate is 105.99 g/mol.
(c) The balanced equation for the titration reaction between sodium tetraborate decahydrate (Na2B4O7•10H2O) and H+ is: Na2B4O7•10H2O + 2H+ + H2O → 4HBO2 + 2Na+ + 12H2O
To find the equivalent weight of sodium tetraborate decahydrate, we need to consider the species that can undergo neutralization. In this case, we have Na2B4O7.
The equivalent weight of a compound is defined as the molecular weight divided by the number of equivalents.
The formula weight of Na2B4O7 can be calculated as follows:
(2*(22.99 g) + 4*(10.81 g) + 7*(16.00 g)) = 381.37 g/mol
The balanced equation shows that 2 moles of H+ react with 1 mole of Na2B4O7, which means there is 1 equivalent of Na2B4O7 per mole.
Therefore, the equivalent weight of sodium tetraborate decahydrate is 381.37 g/mol.
(d) The balanced equation for the titration reaction between Na3PO4•12H2O and HPO42- is: Na3PO4•12H2O + HPO42- → H2PO4- + 3Na+ + 12H2O
To find the equivalent weight of Na3PO4•12H2O, we need to consider the species that can undergo neutralization. In this case, we have Na3PO4.
The equivalent weight of a compound is defined as the molecular weight divided by the number of equivalents.
The formula weight of Na3PO4 can be calculated as follows:
(3*(22.99 g) + 31.00 g + 4*(16.00 g)) = 380.00 g/mol
The balanced equation shows that 1 mole of HPO42- reacts with 1 mole of Na3PO4, which means there is 1 equivalent of Na3PO4 per mole.
Therefore, the equivalent weight of Na3PO4•12H2O is 380.00 g/mol.
To determine the equivalent weight, we need to understand the concept of equivalent weight first. Equivalent weight is the molar mass of a substance divided by the number of equivalents of the specific chemical species involved in a reaction.
(a) Potassium tetraoxalate, KHC2O4•H2C2O4•2H2O:
To find the equivalent weight, we need to determine the number of equivalents involved in the reaction. In this case, the reaction involves the complete neutralization of Potassium tetraoxalate.
The compound KHC2O4•H2C2O4•2H2O contains two acid equivalents, which are H2C2O4 and H2O, and one base equivalent, which is K.
The molar mass of KHC2O4•H2C2O4•2H2O = (39.10 + 1.01 + 12.01 + 4 * 16.00) + (2 * (1.01 + 12.01 + 4 * 16.00)) + 2 * (2 * 1.01 + 16.00) = 404.43 g/mol
Since there is one base equivalent (K) and two acid equivalents (H2C2O4 and H2O), the equivalent weight is equal to the molar mass divided by three:
Equivalent weight = 404.43 g/mol / 3 = 134.81 g/equivalent
Therefore, the equivalent weight of Potassium tetraoxalate is 134.81 g/equivalent.
(b) Sodium carbonate, Na2CO3:
In this case, the reaction is between Sodium carbonate and HCl. The reaction equation is Na2CO3 + HCl = NaCl + NaHCO3.
From the reaction equation, we can see that one mol of Sodium carbonate reacts with one mol of HCl to form one mol of Sodium bicarbonate (NaHCO3) and one mol of Sodium chloride (NaCl).
The molar mass of Na2CO3 = 2 * (23.00) + 12.01 + 3 * 16.00 = 106.00 g/mol
Since one mol of Na2CO3 reacts with one mol of HCl, the equivalent weight is equal to the molar mass:
Equivalent weight = 106.00 g/mol
Therefore, the equivalent weight of Sodium carbonate is 106.00 g/equivalent.
(c) Sodium tetraborate decahydrate, Na2B4O7•10H2O:
In this case, the reaction equation is Na2B4O7•10H2O + 2H+ + H2O = 4HBO2 + 2Na+.
From the reaction equation, we can see that one mol of Sodium tetraborate decahydrate reacts with two mols of H+ to form four mols of Orthoboric acid (HBO2) and two mols of Sodium ion (Na+).
The molar mass of Na2B4O7•10H2O = 2 * (23.00) + 4 * (10.81 + 16.00 + 7 * (16.00)) + 10 * (2 * 1.01 + 16.00) = 381.40 g/mol
Since one mol of Na2B4O7•10H2O reacts with two mols of H+, the equivalent weight is equal to the molar mass divided by two:
Equivalent weight = 381.40 g/mol / 2 = 190.70 g/equivalent
Therefore, the equivalent weight of Sodium tetraborate decahydrate is 190.70 g/equivalent.
(d) Na3PO4•12H2O:
In this case, the reaction is not directly mentioned. However, the question mentions the formation of HPO42- as the product. HPO42- is formed by the partial neutralization of phosphoric acid (H3PO4) by Sodium hydroxide (NaOH).
The molar mass of Na3PO4•12H2O = 3 * (23.00) + 31.00 + 4 * (16.00) + 12 * (2 * 1.01 + 16.00) = 380.14 g/mol
To find the equivalent weight, we need to consider the stoichiometry of the reaction. Since the equation for the reaction is not provided, we cannot determine the exact number of equivalents involved.
We can only make an assumption based on the formation of HPO42-. Assuming one mol of Na3PO4•12H2O reacts with three mols of NaOH to form one mol of HPO42- and three mols of Sodium ions (Na+), we can determine the equivalent weight.
Since one mol of Na3PO4•12H2O reacts with three mols of NaOH, the equivalent weight is equal to the molar mass divided by three:
Equivalent weight = 380.14 g/mol / 3 = 126.71 g/equivalent
Therefore, the equivalent weight of Na3PO4•12H2O is approximately 126.71 g/equivalent.