MCAT Physical Assistance » General Chemisattempt » Acid-Base Chemisattempt » Reactions and also Titrations » Henderson-Hasselbalch Equation

A solution of acetic acid (pKa = 4.75) has a pH of 6.75. The proportion of acid to conjugate base is __________.

You are watching: Use the henderson-hasselbalch equation to calculate the ph of each solution

Explanation:

Use the Henderson-Hasselbalch equation:

}left < HA est >" align="absmiddle">

}left < HA ideal >" align="absmiddle">

}left < HA ideal >" align="absmiddle">

}left < HA est >=100" align="absmiddle">

We want the proportion of acid to conjugate base, which would be the reciprocal,

}left < A^- ight >=frac1100" align="absmiddle">

NaOH is included to a 500mL of 2M acetic acid. If the pKa worth of acetic acid is roughly 4.8, what volume of 2M NaOH need to be included so that the pH of the solution is 4.8?

Explanation:

To resolve this question you should think around the chemical reactivity arising.

We can ignore water and sodium ions for the sake of this question. The reactants exist in a 1:1 proportion, so that for eextremely mol of NaOH we include, we shed one mol of acetic acid and get one mol of acetate. We can identify the moles of acetic acid by using M = mol/L, which gives us mol = ML = (2M) * (0.5L) = 1mol acetic acid. If we use the Hendersen Hasselbach equation we deserve to see that the pH amounts to the pKa as soon as the concentration of conjugate base (acetate) equals the concentration of acid.

See more: Why Was West Berlin Described As A “Bone In The Throat” Of The Soviet Union? ?

If we have 1mol of acetic acid and also include 0.5mol of NaOH, we will certainly lose 0.5mol of acetic acid and get 0.5mol of acetate. We will then be at a point where acetic acid amounts to acetate. This is summarized in the ICE table below. Now we recognize the moles of NaOH (0.5 moles) and the concentration (2M) so we deserve to find the volume by doing M = mol/L.