   Chapter 16, Problem 16.42QP ### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343

#### Solutions

Chapter
Section ### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343
Textbook Problem

# A chemist wanted to determine the concentration of a solution of lactic acid, HC3H5O3. She found that the pH of the solution was 2.60. What was the concentration of the solution? The Kd of lactic acid is 1.4 × 10−4.

Interpretation Introduction

Interpretation:

The concentration of the solution of lactic acid with pH of 2.60 has to be calculated

Concept Information:

Acid ionization constant Ka :

The ionization of a weak acid HA can be given as follows,

HA(aq)H+(aq)+A-(aq)

The equilibrium expression for the above reaction is given below.

Ka=[H+][A-][HA]

Where,

Ka is acid ionization constant,

[H+]   is concentration of hydrogen ion

[A-]   is concentration of acid anion

[HA] is concentration of the acid

pH definition:

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

pH=-log[H+]

On rearranging, we get

[H+]=10pH

To Calculate: The concentration of the solution of lactic acid with pH of 2.60

Explanation

Given data:

The pH of the lactic acid solution = 2.60

The Ka value for the lactic acid = 1.4×104

Calculation of hydrogen ion concentration and lactate ion concentration:

The hydrogen ion concentration can be calculated from the given pH as follows,

[H+] =10pH =102.60 =2.51×103M

Therefore, the concentration of hydrogen ion is 2.51×103M

Represent, lactic acid as HLac and its conjugate anion as Lac

Lactic acid dissociates into hydronium ion (H3O+) and lactate ion (Lac-) .  The hydronium ion concentration equals the concentration of lactate ion.

Therefore, the concentration of lactate ion (Lac-) is 2.51×103M

Calculation of concentration of lactic acid:

Set up the equilibrium table for the given acid solution.

Let x be the unknown concentration.

 HLac(aq)+H2O(l)     ⇄        H3O+(aq)  +   Lac-(aq) Initial (M) x −2.51×10−3M x−2.51×10−3M 0.00 0.00 Change (M) +2.51×10−3M +2

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