To calculate the pH of this solution, we use the
Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Where,
[A-] = Molarity of the conjugate base =
CH3COO- = 0.29 M<span>
<span>[HA] = Molarity of the weak acid = CH3COOH = 0.18 M</span></span>
pKa = dissociation constant of the weak acid =
4.75
When KOH is added to the buffer, the chemical
reaction is:
CH3COOH + KOH = CH3COO-K+ + H2O
Therefore when 0.0090 mol KOH is added, 0.0090
mol acid is neutralized, and 0.0090 mol CH3COO- is produced.
[CH3COO-] = [0.0090 mol + 0.375 L (0.29 mol/L) ]
/ 0.375 L = 0.314 M
[CH3COOH] = [-0.0090 mol + 0.375 L (0.18 mol/L) ]
/ 0.375 L = 0.156 M
Going back to Henderson-Hasselbalch
equation:
pH = 4.75 + log (0.314 / 0.156)
<span>pH = 5.054</span>
Answer:
5.46
Explanation:
From the question,
pH this can be defined as the acidity or alkalinity of a solution.
The expression for pH is given as
pH = -log(H⁺)...................... Equation 1
Where H⁺ = Hydrogen ion of HCL
Given: H⁺ = 3.5×10⁻⁶ M
Substitite this value into equation 1
pH = -log(3.5×10⁻⁶)
pH = 5.46
Hence the pH of HCl is 5.46
Answer:
answering this for points! Sorry! but its probably A OR C
Explanation:
Answer:
naoh is called sodium hydroxide,
Explanation:
hope this helps
