Assuming complete dissociation, calculate the pH of the following solutions:
A. 0.003 M HCl B. 0.005 M NaOH
C. 0.002 M HBr D. 0.002 M KOH
To calculate the pH, we apply the formula:
For acidic solution, pH = -log[H3O+]
For basic solution, pOH = - log[OH-]
Also, we use the given formula to calculate pH and pOH
pH + pOH = 14
A)0.003M HCl
Given:
[HCl] = 0.003 M
Ionization of HCl
HCl + H2O ⇌ H3O+ + Cl-
Since HCl is completely ionized(given)
[HCl] = [H3O+]
⇒[H3O+] = 0.003 M
By applying the formula,
pH = -log[H3O+]
As [H3O+] = 0.003 M
∴ pH = -log[0.003]
⇒pH = 2.52
Thus, the pH of the solution is 2.52
B) 0.005 M NaOH
Given:
[NaOH] = 0.005M
Ionization of NaOH,
NaOH ⇌ Na+ + OH-
Since NaOH is completely ionized (given)
[NaOH] = [OH-]
⇒[OH-] = 0.005 M
By applying the formula for basic solution:
pOH = -log[OH-]
As [OH-] = 0.005 M
∴ pOH = -log [0.005]
⇒pOH = 2.30
As pOH + pH =14
⇒pH = 14 - pOH
⇒pH = 14 – 2.30
⇒pH = 11.70
Thus, the pH of the solution is 11.70
c)0.002 M HBr
Given:
[HBr] = 0.002 M
Ionization of HBr
H2O + HBr ⇌ H3O+ + Br-
Since HBr is completely ionized(given)
[HBr] = [H3O+]
⇒[H3O+] = 0.002 M
By applying the formula (for acidic solution)
pH = -log[H3O+]
As [H3O+] = 0.002 M
∴ pH = -log[0.002]
⇒pH = 2.69
Thus, the pH of the solution is 2.69
d)0.002 M KOH
Given:
[KOH] = 0.002M
Ionization of KOH
KOH + ⇌ K+ + OH-
Since KOH is completely ionized (given)
[KOH] = [OH-]
⇒[OH-] = 0.002 M
By applying the formula for basic solution,
pOH = -log[OH-]
As [OH-] = 0.002 M
∴ pOH = -log[0.002]
⇒pOH = 2.69
As pOH + pH =14
⇒pH = 14 - pOH
⇒pH = 14 – 2.69
⇒pH = 11.31
Thus, the pH of the solution is 11.31