A solution contains 0.250 M HA (K_a = 1.0 $\times$ 10^-⁶)and 0.45 M NaA.What is the pH after 0.17 mole of HCl is added to 1.00 L of this solution?

Question

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To solve the problem, we need to use the Henderson-Hasselbalch equation: pH = pKa + log([A]/[HA]) where pKa is the acid dissociation constant of the weak acid (HA), [A] is the concentration of the conjugate base (A-), and [HA] is the concentration of the weak acid (HA). First, we need to find the concentrations [A] and [HA]: [A] = 0.45 M (given) [HA] = 0.250 M - (moles of HCl added / total volume = 0.17 / 1.00 = 0.17 M) [A-] / [HA] = 0.45 / 0.17 = 2.65 Next, we need to find the pKa of HA. We can use the equation: pKa = -log(Ka) where Ka is the acid dissociation constant. K_a = 1.0 F1F1F1S1F1F1F10 10^-⁶ Assuming the acid is a monoprotic weak acid, the Ka is: Ka = [H+][A-]/[HA] = x^2 / (0.250- x) = 1.6 x 10^-5 Solving for x, we get: x = [H+] = 1.27 x 10^-3 M pKa = -log(1.6 x 10^-5) = 4.80 Finally, we can plug in the values into the Henderson-Hasselbalch equation: pH = 4.80 + log(2.65) = 5.82 Therefore, the pH after 0.17 mole of HCl is added to 1.00 L of this solution is 5.82. The answer is (C).

A Solution Contains 0 ×\times× 10-6)and 0

A Solution Contains 0 $\times$ 10^-⁶)and 0