A dilute tetraoxosulphate(vi)acid B, contains 2.0g of NaOH per 250cm3 of solution. 25cm3 of B requires 20.0cm3 of A for complete neutralization. Calculate:
a. Concentration of B in mol/dm3
b. Concentration of A in mol/dm3
c. Concentration of hydrogen ion in 0.1dm3 of solution
Please tell me the source of your information for naming H2SO4 the name of
tetraoxosulphate(vi)acid. Was it a teacher, your text, a friend,?? That is not a correct name acceptable to IUPAC. Sulfuric acid is an acceptable name and there are others. Tell me that and I'll get back to the answering your question.
a. Concentration of B in mol/dm3:
To find the concentration of B, we need to calculate the number of moles of NaOH in 2.0g.
First, let's convert 250cm3 to dm3:
250cm3 = 0.25dm3
The molar mass of NaOH is:
23 + 16 + 1 = 40g/mol
So, the number of moles of NaOH in 2.0g is:
2.0g / 40g/mol = 0.05mol
Now, we can calculate the concentration of B:
Concentration = Moles / Volume
Concentration = 0.05mol / 0.25dm3 = 0.2mol/dm3
b. Concentration of A in mol/dm3:
Since 25cm3 of B requires 20.0cm3 of A for complete neutralization, we can assume that the concentration of A is the same as B. Therefore, the concentration of A is also 0.2mol/dm3.
c. Concentration of hydrogen ion in 0.1dm3 of solution:
Since 25cm3 of B requires 20.0cm3 of A for complete neutralization, it means that the ratio of the number of moles of hydrogen ions in B (which comes from dilute tetraoxosulphate(VI) acid) and the number of moles of hydrogen ions in A (which comes from the 0.1dm3 solution) is 25:20.
So, the concentration of hydrogen ion in 0.1dm3 of solution can be calculated using this ratio:
Concentration of hydrogen ion = Concentration of A x (Volume of A / Volume of B)
Concentration of hydrogen ion = 0.2mol/dm3 x (20.0cm3 / 25.0cm3)
Concentration of hydrogen ion = 0.16mol/dm3
Therefore, the concentration of hydrogen ion in 0.1dm3 of solution is 0.16mol/dm3.
To calculate the concentrations, we need to follow these steps:
a. Concentration of B in mol/dm3:
Step 1: Calculate the number of moles of NaOH in 25cm3 of B.
The molar mass of NaOH (sodium hydroxide) is 40g/mol.
Number of moles = mass / molar mass
Number of moles of NaOH = 2.0g / 40g/mol = 0.05 mol
Step 2: Calculate the concentration of B in mol/dm3.
Concentration (mol/dm3) = Number of moles / Volume (dm3)
Concentration of B = 0.05 mol / (25 cm3 / 1000) = 2.0 mol/dm3
b. Concentration of A in mol/dm3:
Step 1: Determine the stoichiometric relationship between A and B.
From the question, we know that 25cm3 of B requires 20.0cm3 of A for complete neutralization. This means that the mole ratio of A to B is 20:25 or 4:5.
Step 2: Calculate the concentration of A based on the mole ratio.
Since the concentration of B is 2.0 mol/dm3, and the mole ratio of A to B is 4:5, we can set up the following equation:
Concentration of A / Concentration of B = 4/5
Concentration of A = (4/5) * 2.0 mol/dm3 = 1.6 mol/dm3
c. Concentration of hydrogen ion in 0.1dm3 of solution:
In a neutralization reaction, every mole of acid (A) reacts with an equal number of moles of base (B) to produce an equal number of moles of water (H2O).
Since the reaction is neutral, the concentration of hydrogen ion (H+) will be equal to the concentration of hydroxide ion (OH-), which is the concentration of NaOH in B.
Therefore, the concentration of hydrogen ion in 0.1dm3 of solution is 2.0 mol/dm3.
To solve this problem, we need to use the concept of molarity (concentration) and the reaction stoichiometry. Here's how you can calculate the answers to each part of the question:
a. Concentration of B in mol/dm3:
First, we need to find the number of moles of NaOH in a given volume of solution (B). The molar mass of NaOH is 22.99 g/mol (Na) + 16.00 g/mol (O) + 1.01 g/mol (H) = 39.99 g/mol.
The given mass of NaOH is 2.0 g. Therefore, the number of moles of NaOH in 250 cm3 of solution can be calculated as follows:
Number of moles of NaOH = Mass / Molar mass
Number of moles of NaOH = 2.0 g / 39.99 g/mol
Number of moles of NaOH = 0.050 mol
Now, we need to convert the volume from cm3 to dm3 (1 dm3 = 1000 cm3):
Volume of solution B in dm3 = 250 cm3 / 1000 cm3/dm3
Volume of solution B in dm3 = 0.250 dm3
Finally, we can calculate the concentration (molarity) of B:
Concentration of B = Number of moles of NaOH / Volume of solution B
Concentration of B = 0.050 mol / 0.250 dm3
Concentration of B = 0.20 mol/dm3
Therefore, the concentration of B is 0.20 mol/dm3.
b. Concentration of A in mol/dm3:
Given that 25 cm3 of B requires 20 cm3 of A for complete neutralization, we can assume that the molar ratio between B and A is 1:1. This means that the number of moles of A is the same as the number of moles of B.
Since we know the volume of solution B (0.250 dm3) and the concentration of B (0.20 mol/dm3), we can calculate the number of moles of A:
Number of moles of A = Concentration of B × Volume of solution B
Number of moles of A = 0.20 mol/dm3 × 0.025 dm3
Number of moles of A = 0.005 mol
Now, we need to convert the volume from cm3 to dm3 (1 dm3 = 1000 cm3):
Volume of solution A in dm3 = 20 cm3 / 1000 cm3/dm3
Volume of solution A in dm3 = 0.020 dm3
Finally, we can calculate the concentration (molarity) of A:
Concentration of A = Number of moles of A / Volume of solution A
Concentration of A = 0.005 mol / 0.020 dm3
Concentration of A = 0.25 mol/dm3
Therefore, the concentration of A is 0.25 mol/dm3.
c. Concentration of hydrogen ion in 0.1 dm3 of solution:
Since the molar ratio between B and A is 1:1, we know that the number of moles of NaOH and the number of moles of hydrogen ion (H+) are the same.
The concentration of hydrogen ion in 0.1 dm3 of solution is equal to the concentration of A, which we calculated to be 0.25 mol/dm3.
Therefore, the concentration of hydrogen ion in 0.1 dm3 of solution is 0.25 mol/dm3.