A 2.5 g sample of NaOH (Mw = 40.00) was dissolved in water to give a
solution of final volume 250 cm3.
(i) With reasons, state whether NaOH is a strong or a weak base. Give the
conjugate acid of NaOH and decide whether this conjugate acid is acid,
alkaline or neutral.
(ii) Calculate the molarity NaOH in the solution.
(iii) Calculate the pH of the solution.
(iv) Calculate the molarity of the resulting solution if 100 cm3 of water is
added to the original 250 cm3 NaOH solution. Find the pH of the new
solution.
(v) 4.5 g of acid HA (Mw = 60.00, pKa = 4.8) is added to 100 cm3 of the
original NaOH solution. Assuming no dilution occurs; calculate the pH
of the resulting solution.
(i) To determine whether NaOH is a strong or weak base, we need to consider its dissociation in water. Strong bases dissociate completely in water, while weak bases only partially dissociate. NaOH is a strong base because it completely dissociates into Na+ ions and OH- ions in water.
The conjugate acid of NaOH is H2O, which is a neutral substance. Water is neither acidic nor alkaline.
(ii) To calculate the molarity of NaOH in the solution, we need to use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
First, we need to convert the mass of NaOH to moles by dividing it by the molar mass:
moles of NaOH = mass of NaOH / molar mass of NaOH
= 2.5 g / 40.00 g/mol
= 0.0625 mol
Next, we need to convert the volume of the solution from cm3 to liters:
volume of solution = 250 cm3 / 1000 cm3/L
= 0.25 L
Now we can calculate the molarity:
Molarity of NaOH = 0.0625 mol / 0.25 L
= 0.25 M
(iii) To calculate the pH of the solution, we need to know the concentration of OH- ions. Since NaOH completely dissociates in water, the concentration of OH- ions is equal to the molarity of NaOH that we calculated in part (ii).
pOH = -log[OH-]
= -log(0.25)
= 0.3979
To convert pOH to pH, we can use the equation:
pH = 14 - pOH
= 14 - 0.3979
= 13.6021
Therefore, the pH of the solution is approximately 13.6.
(iv) If 100 cm3 of water is added to the original 250 cm3 NaOH solution, the volume of the resulting solution becomes 250 cm3 + 100 cm3 = 350 cm3.
To calculate the molarity of the resulting solution, we can use the same formula as in part (ii):
Molarity of NaOH = moles of NaOH / volume of solution
Since the moles of NaOH remains the same, and the volume of the solution increases with the addition of water, the molarity will decrease.
Molarity of the new solution = 0.0625 mol / 0.35 L
= 0.1786 M
To find the pH of the new solution, we can follow the same process as in part (iii) using the molarity of the new solution.
(v) To calculate the pH of the resulting solution when 4.5 g of acid HA is added, we need to consider the acid-base reaction between HA and NaOH. It is given that HA has a pKa value of 4.8, which means it will partially dissociate in water, forming H+ ions.
Without knowing additional information about the reaction between HA and NaOH or the Ka value of HA, it is not possible to accurately determine the resulting pH. However, we can make some general observations.
Since NaOH is a strong base and HA is a weak acid, the reaction will favor the formation of water and the corresponding salt. The OH- ions from NaOH will neutralize some of the H+ ions from HA, resulting in a decrease in the overall acidity of the solution. As a result, the pH of the resulting solution is expected to increase compared to the original NaOH solution. However, without more information, a precise pH calculation is not possible.