Answer:
Step-by-step explanation:
In this problem, a 10 - 10 M solution of HCl contributes 10 - 10 M [H +]. The ionization of water contributes 10 - 7 M [H +].
4
HCl ⟶ H⁺ + Cl⁻
HCl dissociates completely in solution, so
1.0 × 10⁻⁴ mol·L⁻¹ HCl ⟶ 1.0 × 10⁻⁴ mol·L⁻¹ H⁺
pH = -log(H⁺) = -log(1.0 × 10⁻⁴) = -log(1.0) - log(10⁻⁴)= -0 - log(-4) = -(-4) = 4
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