Calculate the ratio of NaF to HF required to create a buffer with pH = 3.90 Ka of HF is 6.8 × 10⁻⁴ Solution: Whenever a problem says that there is a buffer, we can use Henderson-Hasselbach equation. NaF is the base and HF is the conjugate acid (has one more hydrogen). Lets plug in the values and solve for the ratio. 3.90 = −log(6.8 × 10⁻⁴) + log([NaF]/[HF]) 3.90 = 3.17 + log([NaF]/[HF]) 0.73 = log([NaF]/[HF]) 10^0.73 = [NaF]/[HF] [NaF]/[HF] = 5.37 LINKS: Chemistry Tutoring Ge

Write an expression for the equilibrium constant of each chemical equation: a. SbCl 5 ( g ) ⇌ SbCl 3 ( g ) + Cl 2 ( g ) b. 2 BrNO( g ) ⇌ 2 NO( g ) + Br 2 ( g ) Solution: To write an expression for equilibrium constant, we do the concentration (symbolized by the brackets) of products to their coefficients over the concentration of reactants to their coefficients. We get: a. K = [SbCl₃][Cl₂] / [SbCl₅] b. K = [NO]²[Br₂] / [BrNO]² LINKS: Chemistry Tutoring General Chemistry Study