Answer :
A 4.85x10⁻³-mol sample of HY is dissolved in enough H₂O to form 0.095 L of solution, if the pH of the solution is 2.68, what is the Ka of HY = 8.84 × 10⁻⁵
What is pH?
The term pH, which originally stood for "potential of hydrogen" (or "power of hydrogen"), is used in chemistry to describe how acidic or basic an aqueous solution is. Lower pH values are summarized for acidic solutions (solutions with higher H+ ion concentrations) than for basic or alkaline solutions.
The pH scale is inversely indicates to the concentration of hydrogen ions in the solution and is logarithmic.
⇒pH = -log([tex]a_{H+}[/tex])
Acidic solutions are those with a pH below 7, and basic solutions are those with a pH above 7, at a temperature of 25 °C (77 °F). At this temperature, solutions with a pH of 7 are neutral (e.g. pure water). The pH neutrality relies on temperature, falling below 7 if the temperature rises above 25 °C.
We have given that HY = 4.85x10⁻³ mol
V = 0.095 L
pH = 2.68
To find concentration of HY
[HY] = mol/V = 4.85x10⁻³ mol/0.095 L
= 0.051 M
We know that pH = -log[H₃O⁺]
pH = -log[H₃O⁺]
[H₃O⁺] = [tex]10^{-pH}[/tex]
= [tex]10^{-2.68}[/tex]
= 2.08× 10⁻³
Lets now make the ICE table to get eq. for Ka
[HY] + H₂O ⇆ H₃O [Y]
I 0 0 0
C -x +x +x
E 0.051 M-x +x +x
Now, we know that H₃O = x = [Y] = 2.08× 10⁻³M
Ka = [tex]\frac{[H_3O^+][Y^-]}{[HY] }[/tex]
= [tex]\frac{x^2}{0.051-x}[/tex]
= [tex]\frac{(2.88*10^{-4})^2}{0.382 - (2.88*10^{-4})}[/tex]
= 8.84 × 10⁻⁵
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