1.76g of sodium hydroxide was added to 50cm^3 of 0.025mol.l^-1 nitric acid. Calculate the pH of the resulting solution.
To calculate the pH of the resulting solution, we first need to determine the moles of sodium hydroxide (NaOH) and nitric acid (HNO3) present.
Step 1: Determine the moles of NaOH.
Given:
- Mass of NaOH = 1.76 g
- Molar mass of NaOH = 23.0 g/mol (sodium: 23.0 g/mol, oxygen: 16.0 g/mol, hydrogen: 1.0 g/mol)
Using the formula:
moles = mass / molar mass
moles of NaOH = 1.76 g / 40.0 g/mol = 0.044 mol
Step 2: Determine the moles of HNO3.
Given:
- Volume of HNO3 solution = 50 cm^3 = 50 mL
- Concentration of HNO3 = 0.025 mol/L
Using the formula:
moles = concentration × volume (in liters)
moles of HNO3 = 0.025 mol/L × 0.050 L = 0.00125 mol
Step 3: Determine the limiting reagent.
The limiting reagent is the reactant that is completely consumed in the reaction or limits the amount of product formed. In this case, NaOH and HNO3 react in a 1:1 molar ratio, so the limiting reagent will be the one with fewer moles.
Since NaOH has 0.044 mol and HNO3 has 0.00125 mol, HNO3 is the limiting reagent.
Step 4: Determine the excess moles of NaOH.
To know the amount of excess NaOH, we subtract the moles of NaOH that reacted with HNO3 from the total moles of NaOH.
Excess moles of NaOH = Total moles of NaOH - Moles of NaOH that reacted
= 0.044 mol - 0.00125 mol
= 0.04275 mol
Step 5: Calculate the concentration of the excess NaOH.
The excess NaOH is present in a volume of 50 mL, which is the same as 0.050 L.
Concentration of excess NaOH = Excess moles of NaOH / Volume (in liters)
= 0.04275 mol / 0.050 L
= 0.855 mol/L
Step 6: Calculate the concentration of OH- ions in the solution.
Since NaOH is a strong base, it dissociates completely in water to form sodium ions (Na+) and hydroxide ions (OH-). Knowing the concentration of excess NaOH, we can determine the concentration of OH- ions.
The concentration of OH- ions is equal to the concentration of excess NaOH because of the 1:1 stoichiometric ratio between NaOH and OH-.
Concentration of OH- ions = 0.855 mol/L
Step 7: Calculate pOH.
The pOH is the negative logarithm (base 10) of the hydroxide ion concentration.
pOH = -log10 (concentration of OH- ions)
= -log10 (0.855)
≈ 0.067
Step 8: Calculate the pH.
The pH is the negative logarithm (base 10) of the hydrogen ion concentration (H+). Since water dissociates into H+ and OH-, the concentrations of H+ and OH- will be equal in a neutral solution.
pH + pOH = 14
pH = 14 - pOH
= 14 - 0.067
≈ 13.933
The pH of the resulting solution is approximately 13.933.
To calculate the pH of the resulting solution, we need to determine the concentration of the resulting solution after the reaction between sodium hydroxide (NaOH) and nitric acid (HNO3).
Step 1: Write the balanced chemical equation for the reaction:
NaOH + HNO3 → NaNO3 + H2O
Step 2: Determine the limiting reagent:
To do this, we need to compare the moles of each reactant.
Given:
Mass of NaOH = 1.76 g
Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
Number of moles of NaOH = Mass / Molar mass = 1.76 g / 40 g/mol = 0.044 mol
Moles of HNO3 = Molarity × Volume
Given:
Molarity of HNO3 = 0.025 mol/L
Volume of HNO3 = 50 cm^3 = 50 mL = 50/1000 L = 0.05 L
Moles of HNO3 = 0.025 mol/L × 0.05 L = 0.00125 mol
Comparing the moles of NaOH and HNO3, we can see that NaOH is in excess because it has a higher number of moles.
Step 3: Calculate the moles of excess NaOH:
Moles of excess NaOH = Moles of NaOH - Moles of HNO3
= 0.044 mol - 0.00125 mol
= 0.04275 mol
Step 4: Calculate the concentration of the resulting solution:
The number of moles of the resulting solution depends on the volume of the solution.
Given that the volume of the solution is 50 mL, or 50/1000 L = 0.05 L.
Concentration of the resulting solution = Moles / Volume
= 0.04275 mol / 0.05 L
= 0.855 mol/L
Step 5: Calculate the pOH of the resulting solution:
pOH = -log10[OH-]
Since NaOH is a strong base, it will completely dissociate, so the concentration of OH- ions in the solution will be the same as the concentration of NaOH.
pOH = -log10(0.855)
= 0.067
Step 6: Calculate the pH of the resulting solution:
pH = 14 - pOH
= 14 - 0.067
= 13.933
Therefore, the pH of the resulting solution is approximately 13.933.
NaOH+HNO3>>NaNO3 + water
moles acid= .050dm^3*.025mol/dm^3
= 1.25mMole (check that).
Moles NaOH=1.76/40= 44mMole (check that).
moles of NaOH in excess:44-1.25=you do it.
This represents the OH concentration...
OH=(molesOHabove/.05)
H= 1E-14/OHabove
pH= -log(H)