a) A suitable indicator for the reaction between trioxonitrate(V) acid (HNO3) and sodium hydroxide (NaOH) would be phenolphthalein. Phenolphthalein is commonly used as an indicator for acid-base titrations because it exhibits a color change from colorless to pink around pH 8.2 to 10.0, which is in the range of the reaction between an acid and a base.
b) The balanced equation for the reaction between trioxonitrate(V) acid (HNO3) and sodium hydroxide (NaOH) is:
HNO3 + NaOH → NaNO3 + H2O
c) To calculate the concentrations, we need to use the given information:
Solution B:
- Mass of NaOH per dm3: 4.00 g
- Molar mass of NaOH: Na = 23, O = 16, H = 1
- Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
- Moles of NaOH in 1 dm3 = mass/Molar mass = 4.00 g / 40 g/mol = 0.10 mol/dm3
1. The concentration of solution B in moles per dm3 is 0.10 mol/dm3.
Solution A does not have enough information provided to directly calculate its concentration in moles per dm3. However, if a titration is performed between solution A and solution B, we can use the volume of solution B required to neutralize solution A to calculate the concentration of solution A in moles per dm3.
3. To calculate the concentration of grams per dm3 of HNO3 in solution A, we need to use the balanced equation from part b and the result from part c(1). From the balanced equation, we know that 1 mole of HNO3 reacts with 1 mole of NaOH.
Therefore, the concentration of grams per dm3 of HNO3 in solution A will be equal to the concentration of solution B in moles per dm3, which is 0.10 mol/dm3. We also need to consider the molar mass of HNO3 to convert moles to grams.
- Molar mass of HNO3: H = 1, N = 14, O = 16
- Molar mass of HNO3 = 1 + 14 + (3 * 16) = 63 g/mol
The concentration of grams per dm3 of HNO3 in solution A is:
0.10 mol/dm3 * 63 g/mol = 6.30 g/dm3
Therefore, the concentration of grams per dm3 of HNO3 in solution A is 6.30 g/dm3.