A is a solution of trioxonitate acid of an unknown concentration, B is a standard solution of sodium hydroxide containing 4.00g per dm3 of solution 25cm3 portion of solution B require an average of 24cm3 of solution A for complete neutralization. Write a balance equation for the reaction.calculate
A. Concentration of B in grame per dm3
B. Concentration of A in grame per dm3
C. Concentration in grame per dm3 of the trioxonitrate(v) acid in solution A
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To solve this problem, we need to use the balanced equation for the reaction between trioxonitrate(V) acid (A) and sodium hydroxide (B). Let's assume the balanced equation is:
HNO3 + NaOH → NaNO3 + H2O
A. To calculate the concentration of B in grams per dm3, we'll need to find the molarity (M) of solution B. Molarity is defined as the number of moles of solute (in this case, NaOH) per liter of solution (dm3).
Given that solution B contains 4.00g per dm3 of NaOH, we need to convert grams to moles using its molar mass. The molar mass of NaOH is calculated by adding the atomic masses of sodium (Na), oxygen (O), and hydrogen (H):
Na: 22.99 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
NaOH = Na + O + H
= 22.99 g/mol + 16.00 g/mol + 1.01 g/mol
= 40.00 g/mol
Now we can calculate the molarity (M) of B:
Molarity (M) = (mass of solute in grams / molar mass of solute) / volume of solution in liters
= (4.00 g / 40.00 g/mol) / 0.025 dm3
= 0.100 mol/dm3
Therefore, the concentration of B in grams per dm3 is 0.100 g/dm3.
B. To find the concentration of A in grams per dm3, we'll use the stoichiometry of the balanced equation. From the equation, we know that 1 mole of NaOH reacts with 1 mole of HNO3.
Since 25 cm3 of B reacts with an average of 24 cm3 of A, we can say that the ratio of the volumes is 25:24. This implies that the moles of B and A are also in a 25:24 ratio.
From our calculation in part A, we know that the molarity (M) of B is 0.100 mol/dm3. Therefore, we can calculate the molarity of A:
Molarity (A) = (Molarity (B) * volume of B) / volume of A
= (0.100 mol/dm3 * 0.025 dm3) / 0.024 dm3
= 0.104 mol/dm3
Now we need to convert the molarity of A to grams per dm3 by multiplying it by the molar mass of HNO3. The molar mass of HNO3 is calculated by adding the atomic masses of hydrogen (H), nitrogen (N), and oxygen (O):
HNO3 = H + N + O
= 1.01 g/mol + 14.01 g/mol + 3(16.00 g/mol)
= 1.01 g/mol + 14.01 g/mol + 48.00 g/mol
= 63.02 g/mol
Concentration of A in grams per dm3 = (Molarity (A) * molar mass of HNO3)
= (0.104 mol/dm3 * 63.02 g/mol)
= 6.53 g/dm3
Therefore, the concentration of A in grams per dm3 is 6.53 g/dm3.
C. The concentration in grams per dm3 of the trioxonitrate(V) acid in solution A is the same as the concentration of A, so it is 6.53 g/dm3.