We are given that the concentration of NaOH is 0.0003 M and are asked to calculate the pH
We know that NaOH dissociates by the following reaction:
NaOH → Na⁺ + OH⁻
Which means that one mole of NaOH produces one mole of OH⁻ ion, which is what we care about since the pH is affected only by the concentration of H⁺ and OH⁻ ions
Now that we know that one mole of NaOH produces one mole of OH⁻, 0.0003M NaOH will produce 0.0003M OH⁻
Concentration of OH⁻ (also written as [OH⁻]) = 3 * 10⁻⁴
<u>pOH of the solution:</u>
pOH = -log[OH⁻] = -log(3 * 10⁻⁴)
pOH = -0.477 + 4
pOH = 3.523
<u>pH of the solution:</u>
We know that the sum of pH and pOH of a solution is 14
pH + pOH = 14
pH + 3.523 = 14 [subtracting 3.523 from both sides]
pH = 10.477
The number of calories that are required to change the temperature of 2.18 g of water from 15.3 c to 69.5 c is <u>118.16 cal</u>
<u><em> calculation</em></u>
- Heat in calories = MCΔ T where,
- M(mass)= 2.18 g
- C(specific heat capacity)= 1.00 cal/g/c
- ΔT( change in temperature)= 69.5- 15.3 =54.2 c
heat is therefore= 2.18 g x 1.00 cal/g/c x 54.2 c=118.16 cal
