A is a solution of trioxonitrate (V) acid. HNO of unknown concentration B is standard solution of sodium hydroxide containing 4.00g per dm3 of solution 25cm3 portion solution B required an average of 24.00cm3 of solution A for neutralization. Using 2 drop of methyl orange.
(a) write a balanced equation for the reaction.
(b) calculate the concentration of solution B in mole per dm3.
(c) calculate the concentration of solution A in moles per dm3.
(d) calculate the concentration in g/dm3 of HNO in solution A. (H=1, N=14, O=16, Na=23)
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(a) The balanced equation for the reaction between the trioxonitrate (V) acid (solution A) and sodium hydroxide (solution B) can be written as:
HNO₃ + NaOH -> NaNO₃ + H₂O
(b) To calculate the concentration of solution B in moles per dm³, we need to determine the number of moles of sodium hydroxide (NaOH) present in the 25 cm³ portion of solution B.
Given:
Mass of NaOH in 1 dm³ of solution B = 4.00 g
Volume of solution B used = 25 cm³
We can convert the volume to dm³ by dividing by 1000:
Volume of solution B used = 25 cm³ ÷ 1000 = 0.025 dm³
Now, we can calculate the concentration of solution B in moles per dm³ (Molarity) using the formula:
Concentration (in mol/dm³) = Moles / Volume (in dm³)
Number of moles of NaOH = Mass / Molar mass
Molar mass (NaOH) = 23 + 16 + 1 = 40 g/mol
Number of moles of NaOH in 0.025 dm³ = 4.00 g / 40 g/mol = 0.10 mol
Concentration of solution B = 0.10 mol / 0.025 dm³ = 4.0 mol/dm³
Therefore, the concentration of solution B is 4.0 mol/dm³.
(c) To calculate the concentration of solution A in moles per dm³, we will use the volume of solution A used for neutralization.
Volume of solution A used = 24.00 cm³
Again, we convert the volume to dm³:
Volume of solution A used = 24.00 cm³ ÷ 1000 = 0.024 dm³
Since the stoichiometry of the balanced equation is 1:1 between HNO₃ and NaOH, the moles of HNO₃ in solution A will be the same as the moles of NaOH in solution B.
Concentration of solution A = Concentration of solution B = 4.0 mol/dm³
Therefore, the concentration of solution A is also 4.0 mol/dm³.
(d) To calculate the concentration of HNO₃ in solution A in g/dm³, we need to use the molar mass of HNO₃.
Molar mass (HNO₃) = (1*1) + (14*1) + (16*3) = 63 g/mol
Concentration of HNO₃ in g/dm³ = Concentration (in mol/dm³) * Molar mass (in g/mol)
Concentration of HNO₃ in g/dm³ = 4.0 mol/dm³ * 63 g/mol = 252 g/dm³
Therefore, the concentration of HNO₃ in solution A is 252 g/dm³.