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Chapter 9: QIE (page 207)

A solution contains 63 different conjugate acid-base pairs.

Among them is acrylic acid and acrylate ion, with the equilibrium ratio [acrylate]/[acrylic acid]5 0.75. What is the pH of the solution?

H2C=CHCO2HpKa=4.25

Short Answer

Expert verified

The pH of the solution using Henderson-Hasselbalch equation and was found to be 4.13

Step by step solution

01

Step 1: Define the term Henderson-Hasselbalch Equation.

Equation of Henderson-Hasselbalch. pH, pKa, and molar concentration (concentration in moles per litre) are all linked by the Henderson-Hasselbalch equation: apH=logA-/HA+pK a The molar concentration of a conjugate base isA-.HA= molar concentration of a weak acid that has not been dissociated (M).

02

Calculate the ph of the solution.

Equilibrium ratio=0.75

pKa=4.25

The pHof the solution can be calculated as,

role="math" localid="1654769713942" pH=pKa+logA-HApH=4.25+log0.75pH=4.13

The pH of the solution=4.13

The pH of the solution using Henderson-Hasselbalch equation and was found to be 4.13.

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Most popular questions from this chapter

A buffer was prepared by dissolving 0.100 mole of the weak acid HA(Ka=1.00脳10-5)plus 0.050 mole of its conjugate baseNa+A-in 1.00 L. Find the pH

Find the pH if only 6.0 instead of 12.0 mL HCl were added.

Neglecting activity coefficients, calculate the pH of5.0脳10-8MHCIO4What fraction of H+is derived from dissociation of water?

Use the Goal Seek spreadsheet at the end of the chapter to find the pH of 1.00 L of solution containing 0.030 molHA(pKa=2.50)and What would the be with the approximations [HA]=0.030 and [A-]=0.015?

Find the pH if only 6.30 instead of 12.0 mL HCl were added
See all solutions

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