The pH of a 0.60 M aqueous solution of formic acid, HCHO₂, at 25.0°C is 1.98. What is the value of K_a for formic acid?

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Formic acid, HCHO₂, is a weak acid and its dissociation in water is represented as follows: HCHO₂(aq) + H2O(l) ⇌ H3O+(aq) + HCOO-(aq) The expression for the acid ionization constant (Ka) is given by: Ka = [H3O+][HCOO-]/[HCHO₂] We can use the pH of the solution to determine the concentration of H3O+ as follows: pH = -log[H3O+] [H3O+] = 10^-pH Substituting the given values, we get: [H3O+] = 10^-1.98 = 1.10 × 10^-2 M Since formic acid is a weak acid, we can assume that its initial concentration is equal to its equilibrium concentration. Let x be the concentration of H3O+ and HCOO- at equilibrium. Then we can write: [H3O+] = [HCOO-] = x [HCHO₂] = 0.60 M - x Substituting these values into the expression for Ka, we get: Ka = x^2/(0.60 - x) Since Ka is a small number, we can assume that x is much smaller than 0.60. This approximation is valid if x << 0.60. Then we can simplify the expression for Ka as follows: Ka = x^2/0.60 Substituting the value of x from the pH calculation, we get: Ka = (1.10 × 10^-2)^2/0.60 = 1.8 × 10^-4 Therefore, the answer is choice B: 1.8 × 10^-4

The PH of a 0