A is a solution of trioxonitrate(v)acid (nitric acid) HNO3 of unknown concentration.B is a standard solution of sodium hydroxide containing 4.00g per dm3 of solution
a). Suggest with a reason,a suitable indicator for the reaction
b) Write a balanced equation for the reason
c) Calculate
1. The concentration of solution B in moles per dm3
2. The concentration of solution A in moles per dm3
3. The concentration of grams per dm3 of HNO3 in solution A (H=1, N=14, O=16, Na=23)
Concentration of B in mol per dm3/ Molar mass= 4.00/40= 0.1
Molar mass of NaOH (23x1)+(16x1)+(1x1)= 40
Conc of solution A Ca VA/Cb Vb=na/nb
Ca? VA 24.00
Cb 0.1mol per dm3 Vb 25cm3
Cax24.00/0.1x25
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, and it changes color from colorless to pink as the pH moves from acidic to slightly basic.
b) The balanced equation for the reaction between HNO3 and NaOH is:
HNO3 + NaOH → NaNO3 + H2O
c)
1. The concentration of solution B in moles per dm3:
The given concentration of solution B is 4.00g per dm3. To calculate the concentration in moles per dm3, we need to know the molar mass of NaOH. The molar mass of NaOH is 23 + 16 + 1 = 40 g/mol.
Concentration of solution B in moles per dm3 = (mass of NaOH / molar mass of NaOH) / volume of solution B
Concentration of solution B = (4.00 g / 40 g/mol) / 1 dm3
Concentration of solution B = 0.10 mol/dm3
2. The concentration of solution A in moles per dm3:
The concentration of solution A is unknown.
3. The concentration of grams per dm3 of HNO3 in solution A:
The molar mass of HNO3 is 1 + 14 + (3 × 16) = 63 g/mol.
Concentration of HNO3 in grams per dm3 = concentration of solution A in moles per dm3 × molar mass of HNO3
Concentration of HNO3 = concentration of solution A × 63 g/dm3
Since the concentration of solution A is unknown, we cannot calculate the concentration of HNO3 in grams per dm3 at the moment.
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.
(HNO3) = ?
a. There are a number of indicators but I would use phenolphthalein (pH change about 8) although there are others that change at pH = 7.0
b. HNO3 + NaOH ==> NaNO3 + H2O
c1. (NaOH) = 4.00 g/dm3.
mols NaOH = 4.00/40 g/mol = 0.1 mol/dm3 = 0.1 M
c2 and c3 can't be done. You didn't give any information about the volume of HNO3 nor the volume of NaOH used.