We can covert 100 mL of NaOH solution with a pH of 13 to a solution with a pH of 12 by adding 900 mL of water to the 100 mL of NaOH solution.
Determination of the pOH of NaOH solution with pH 13
pH + pOH = 14
13 + pOH = 14
pOH = 14 – 13
pOH = 1
Determination of the molarity of NaOH solution with pH 13 (i.e pOH = 1)
[OH¯] = 0.1 M
Finally, we shall determine the molarity of the NaOH solution.
NaOH(aq) —> Na⁺(aq) + OH¯(aq)
0.1 M NaOH will also contain 0.1 M OH¯
pOH = 2
Determination of the molarity of NaOH solution with pH 12 (i.e pOH = 2)
We'll begin by calculating the concentration of the hydroxide ion [OH¯]
pOH = 2
Concentration of the hydroxide ion [OH¯] =?
pOH = –Log [OH¯]
2 = –Log [OH¯]
[OH¯] = 0.01 M
Finally, we shall determine the molarity of the NaOH solution.
NaOH(aq) —> Na⁺(aq) + OH¯(aq)
From the balanced equation above,
1 mole of NaOH contains 1 mole of OH¯.
0.01 M NaOH will also contain 0.01 M OH¯
Molarity of diluted solution (M₂) = 0.01
M₁V₁ = M₂V₂
10 = 0.01 × V₂
Divide both side by 0.01
V₂ = 1000 mL
Determination of the volume of water needed
Volume of water = 1000 – 100
Volume of water needed = 900 mL
Thus, to change the pH of 100 mL of NaOH solution from pH of 13 to pH of 12, added 900 mL of water to the initial solution.
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