Question

O ACIDS AND BASES Calculating the pH of a strong base solution A chemist dissolves 287. mg of pure sodium hydroxide in enough water to make up 80. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Round your answer to 2 significant decimal places. X 5 ?

Answer 1

Molar mass of NaOH,
MM = 1*MM(Na) + 1*MM(O) + 1*MM(H)
= 1*22.99 + 1*16.0 + 1*1.008
= 39.998 g/mol


mass(NaOH)= 287 mg
= 0.287 g

use:
number of mol of NaOH,
n = mass of NaOH/molar mass of NaOH
=(0.287 g)/(40 g/mol)
= 7.175*10^-3 mol
volume , V = 80 mL
= 8*10^-2 L


use:
Molarity,
M = number of mol / volume in L
= 7.175*10^-3/8*10^-2
= 8.969*10^-2 M

So,
[OH-] = 8.969*10^-2 M

use:
pOH = -log [OH-]
= -log (8.969*10^-2)
= 1.0473


use:
PH = 14 - pOH
= 14 - 1.0473
= 12.9527
Answer: 12.95