The pH of a solution is 9.02.
c(HCN) = 1.25 M; concentration of the cyanide acid
n(NaCN) = 1.37 mol; amount of the salt
V = 1.699 l; volume of the solution
c(NaCN) = 1.37 mol ÷ 1.699 l
c(NaCN) = 0.806 M; concentration of the salt
Ka = 6.2 × 10⁻¹⁰; acid constant
pKa = -logKa
pKa = - log (6.2 × 10⁻¹⁰)
pKa = 9.21
Henderson–Hasselbalch equation for the buffer solution:
pH = pKa + log(cs/ck)
pH = pKa + log(cs/ck)
pH = 9.21 + log (0.806M/1.25M)
pH = 9.21 - 0.19
pH = 9.02; potential of hydrogen
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Answer:
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Explanation:
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Answer:
[H₃O⁺] = 2.5 × 10⁻¹⁰ M
pH = 9.6
Explanation:
Step 1: Given data
Concentration of OH⁻ in the solution ([OH⁻]): 4.0 × 10⁻⁵ M
Step 2: Calculate the concentration of H₃O⁺ in the solution
Let's consider the self-ionization of water.
2 H₂O(l) ⇄ H₃O⁺(aq) + OH⁻(aq)
The <em>ion-product of water (Kw)</em> is:
<em>Kw = 1.0 × 10⁻¹⁴ = [H₃O⁺] × [OH⁻]</em>
[H₃O⁺] = 1.0 × 10⁻¹⁴/[OH⁻]
[H₃O⁺] = 1.0 × 10⁻¹⁴/4.0 × 10⁻⁵
[H₃O⁺] = 2.5 × 10⁻¹⁰ M
Step 3: Calculate the pH of the solution
We will use the following expression.
<em>pH = -log [H₃O⁺]</em>
pH = -log 2.5 × 10⁻¹⁰
pH = 9.6