Answer:
A. 400g of NaOH.
B. 0.253g of NaOH.
Explanation:
A. The following data were obtained from the question:
Volume = 5 L
Molarity = 2 M
Mass =..?
Next, we shall determine the number of mole of NaOH. This can be obtain as follow:
Molarity = mole /Volume
2 = mole/5
Cross multiply
Mole = 2 x 5
Mole of NaOH = 10 moles
Finally, we shall convert 10 moles of NaOH to grams. This is illustrated below:
Mole of NaOH = 10 moles
Molar mass of NaOH = 23 + 16 + 1 = 40g/mol
Mass of NaOH =?
Mole = mass /molar mass
10 = mass of NaOH /40
Cross multiply
Mass of NaOH = 10 x 40
Mass of NaOH = 400g
Therefore, 400g of NaOH is needed to prepare the solution.
B. The following data were obtained from the question:
pH = 11.5
Volume = 2 L
Next, we shall determine the pOH of the solution. This can be obtain as shown below:
pH + pOH = 14
pH = 11.5
11.5 + pOH = 14
Collect like terms
pOH = 14 – 11.5
pOH = 2.5
Next, we shall determine the concentration of the hydroxide ion, [OH-] in the solution.
This is illustrated below:
pOH = - Log [OH-]
pOH = 2.5
2.5 = - Log [OH-]
-2.5 = Log [OH-]
Take the antilog of both side
[OH-] = antilog (-2.5)
[OH-] = 3.16×10¯³ M
Next, we shall determine the concentration of NaOH. This is illustrated below:
NaOH —> Na+ + OH-
From the balanced equation above,
1 mole of NaOH produced 1 mole of OH-.
Therefore, 3.16×10¯³ M NaOH will also produce 3.16×10¯³ M OH-.
Therefore, the concentration of NaOH is 3.16×10¯³ M
Next, we shall determine the number of mole of NaOH in the solution. This can be obtain as follow:
Molarity = 3.16×10¯³ M
Volume = 2 L
Mole of NaOH =?
Molarity = mole /Volume
3.16×10¯³ = mole of NaOH / 2
Cross multiply
Mole of NaOH = 3.16×10¯³ x 2
Mole of NaOH = 6.32×10¯³ mole
Finally, we shall convert 6.32×10¯³ mole of NaOH to grams
Mole of NaOH = 6.32×10¯³ mole
Molar mass of NaOH = 23 + 16 + 1 = 40g/mol
Mass of NaOH =?
Mole = mass /molar mass
6.32×10¯³ = mass of NaOH /40
Cross multiply
Mass of NaOH = 6.32×10¯³ x 40
Mass of NaOH =0.253 g
Therefore, 0.253g of NaOH is needed to prepare the solution.
