# Question Video: Calculating the Equilibrium Constant of the Weak Acid HPO₄²⁻

Consider the equilibrium of the HPO₄²⁻ ion acting as a weak base. HPO₄²⁻(aq) + H₂O(l) ⇌ H₂PO₄⁻(aq) + OH⁻(aq), What is the equilibrium constant for this reaction under these conditions? [OH⁻] = 1.3 × 10⁻⁶ M, [H₂PO₄⁻] = 0.042 M, [HPO₄²⁻] = 0.341 M.

03:20

### Video Transcript

Consider the equilibrium of the HPO₄²⁻ ion acting as a weak base. HPO₄²⁻ aqueous plus H₂O liquid in equilibrium with H₂PO₄⁻ aqueous plus OH− aqueous. What is the equilibrium constant for this reaction under these conditions? Concentration of our each minus equal to 1.3 times 10 to the minus six molars, concentration of H₂PO₄⁻ equal to 0.042 molars, and concentration of HPO₄²⁻ equal to 0.341 molars.

Before we go any further, it might be helpful to understand the names for a couple of these ions. Hydrogen phosphate and dihydrogen phosphate all have at their core the phosphate anion, PO₄³⁻. You can get the full name by just adding the number of hydrogens to the beginning: hydrogen for one or dihydrogen for two. In this equilibrium, hydrogen phosphate is acting as a weak base, taking a proton from a water molecule to form dihydrogen phosphate and hydroxide. our job is to figure out the equilibrium constant 𝑘 𝑐 for this reaction, where the concentrations of the components are specified.

Since here hydrogen phosphate is acting as a base, we can also describe this equilibrium as 𝑘 𝑏 of the hydrogen phosphate anion. For the equilibrium where chemicals 𝐴 and 𝐵 are in equilibrium with 𝐶 and 𝐷, this is the equilibrium constant expression. The presence of any pure liquids or pure solids is ignored. Only gases or solutes will appear in the equilibrium constant expression.

We start constructing our equilibrium constant expression by taking the concentrations of our products and multiplying them together. All the components of this equilibrium are in a one-to-one ratio with one another. So we don’t have any funny powers. We finished the expression by adding in the concentration of the hydrogen phosphate anion at the bottom, ignoring the concentration of liquid water. Now all that remains is to substitute in our concentration values, giving us 0.042 molars times 1.3 times 10 to the minus six molars all over 0.341 molars. This evaluates to 1.60117 times 10 to the minus seven molars.

The least significant value in our calculation was given to two significant figures. So we should give the answer to the same precision, meaning our final answer is 1.6 times 10 to the minus seven molars. However, equilibrium constants are conventionally given without units. So our final answer is simply 1.6 times 10 to the minus seven.
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}