The correct answer to this question is that the length of 14 is it’s half Which would be 7
27/208 = normality
12 x 10^-2 approx = normality
nw Ka = 14.3 x 10^-3
pKa = 3 - log 14
now, after getting the pKa put it in formula :
pH = pKa + log concn of ion/concn of salt and you'll get it
hope this helps
Answer:
I didn't do the observation so I can't help sorry
We need to first find the molarity of Ba(OH₂) solution.
A mass of 3.24 mg is dissolved in 1 L solution.
Ba(OH)₂ moles dissolved - 3.24 x 10⁻³ g/171.3 g/mol = 1.90 x 10⁻⁵ mol
dissociaton of Ba(OH)₂ is as follows;
Ba(OH)₂ --> Ba²⁺ + 2OH⁻
1 mol of Ba(OH)₂ dissociates to form 2OH⁻ ions.
Therefore [OH⁻] = (1.90 x 10⁻⁵)x2 = 3.8 x 10⁻⁵ M
pOH = -log[OH⁻]
pOH = -log (3.8 x 10⁻⁵)
pOH = 4.42
pH + pOH = 14
therefore pH = 14 - 4.42
pH = 9.58
