Question

Calculate the pH of the following solutions (1) 0.02mhcl. (2) 0.10mnaoh. (3) 0.0032mhno. (4) 0.03mkoh​

Answer 1

Answer:

1. 1.70

2. 13.00

3. 2.49

4. 12.48

Explanation:

Background information and equations:

• For strong acids, pH can be calculated by $pH = -\log[H_3O^+] = -\log[HX]$,
• For strong bases, pOH is calculated similarly by $pOH = -\log[OH^-] = -\log[XOH]$, at room temperature, $pH = 14.00 - pOH = 14.00 + \log[XOH]$.

1. HCl is a strong acid, given its molarity:

$pH = -\log(0.02) = 1.70$

2. NaOH is a strong base:

$pH = 14.00 + \log(0.10) = 13.00$

3. Nitric acid is a strong acid:

$pH = -log(0.0032) = 2.49$

4. KOH is a strong base:

$pH = 14.00 + log(0.03) = 12.48$

Answer 2

Answer:

1) pH of HCl = 1.7

2) pH of NaOH = 13

3) pH of HNO₃ = 2.5

4) pH of KOH = 12.5

Explanation:

answer is in attachment