Chapter 9: Q37P (page 209)

Calculate the quotient $[{HCO}_{2}^{-}] / [{HCO}_{2} H]$ at $pH 3.744$ if the ionic strength is $0.1 M$ by using the effective equilibrium constant listed for $μ = 0.1$ in Appendix G.

Short Answer

Expert verified

The quotient is $[{HCO}_{2}^{-}] / [{HCO}_{2} H] = 1.493$ .

Step by step solution

Concept used.

The pH of the buffer determined by the concentration ratio of $[A^{-}] and [HA]$ and by the following Henderson-Hasselbalch equation: $pH = {pK}_{a} + \log ([A^{-}] / [HA])$

The quotient.

Determine the ${pK}_{a}$ at $μ = 0.1$ which is $3.57$ and then calculate the following quotient:

$3.744 = 3.570 + \log ([H C O_{2}^{-}]) / [H C O_{2} H] 0.174 = \log ([H C O_{2}^{-}]) / [H C O_{2} H] 10^{0.174} = ([H C O_{2}^{-}]) / [H C O_{2} H] 1.493 = ([H C O_{2}^{-}]) / [H C O_{2} H]$

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Calculate the quotient $[{HCO}_{2}^{-}] / [{HCO}_{2} H]$ at $pH 3.744$ if the ionic strength is $0.1 M$ by using the effective equilibrium constant listed for $μ = 0.1$ in Appendix G.

Short Answer

Step by step solution

Concept used.

The quotient.

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Calculate the quotient [HCO2-]/[HCO2H]at pH3.744if the ionic strength is 0.1Mby using the effective equilibrium constant listed forμ=0.1in Appendix G.

Short Answer

Step by step solution

Concept used.

The quotient.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Chemistry Branches

Inorganic Chemistry

Organic Chemistry

Nuclear Chemistry

Chemical Reactions

Study anywhere. Anytime. Across all devices.

Calculate the quotient $[{HCO}_{2}^{-}] / [{HCO}_{2} H]$ at $pH 3.744$ if the ionic strength is $0.1 M$ by using the effective equilibrium constant listed for $μ = 0.1$ in Appendix G.