pH, pOH, H30+, OH-, Kw, Kb, Ka, pKa, pKb

How to solve equations when given certain information.

How would I calculate the pKa and pKb of an acid when I’m given the Ka value?

Since I am given the Ka value, I would want to solve for pKa first. To do this, I would use the equation:

• pKa = -logKa

Once I find my pKa value, I can then solve for pKb by using the equation:

• 14.00 = pKa + pKb

How would I solve for [OH^-] if I am given [H₃O⁺]?

There are two ways to solve for this.

First:

• Since I am given the [H₃O⁺], I can first solve for pH using this equation

  • pH = -log [H₃O⁺]

• Once I solve for pH, I would then solve for pOH using this equation

  • 14.00 = pH + pOH

• Once I find pOH, I can then solve for [OH^-] using this equation

  • pOH = -log [OH^-]

Second:

• Since I am given the [H₃O⁺], I can use the equation

  • Kw = [H₃O⁺] [OH^-]

  • The Kw = 1.0 × 10^-14

  • I would just rearrange the equation to solve for [OH^-]

How would I solve for the Kb value if I am given the pKa?

There are two ways to solve this problem.

First:

• Since I am given the pKa value, I would solve for Ka using this equation

  • pKa = -logKa

• Once I find the Ka, I can then solve for Kb by using the equation

  • Kw = (Ka)(Kb)

  • Kw = 1.0 × 10^-14

Second:

• Since I am given pKa, I can solve for pKb by using the equation

  • 14.00 = pKa + pKb

• Once I find my pKb value, I can solve for Kb by using the equation

  • pKb = -log [OH^-]

How would I solve for pH of a solution if I am given the [OH^-]?

Since I am given the [OH^-], I would solve for pOH using the equation

• pOH = -log [OH^-]

Once I have my pOH value, I can solve for pH by using the equation

• 14.00 = pH + pOH