Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Calculate pH of \(0.02 \mathrm{M}-\) HA solution. \(K_{\mathrm{a}}\) for \(\mathrm{HA}=2 \times 10^{-12} .(\log 2=0.3\) \(\log 3=0.48\) ) (a) \(6.65\) (b) \(6.70\) (c) \(6.85\) (d) \(6.52\)

Short Answer

Expert verified
6.70

Step by step solution

01

Understanding pH and Acid Dissociation

The pH of a solution is a measure of its acidity or basicity. For a weak acid HA dissociating as HA → H+ + A-, the pH is related to the acid dissociation constant (Ka) and the concentration of the acid ([HA]). The relevant equation for a weak mono-protic acid is: Ka = [H+][A-]/[HA].
02

Set Up the Equilibrium Equation

Assume that the acid dissociates partially, and x is the concentration of H+ and A- at equilibrium. Since the dissociation is a 1:1 ratio, [H+] = [A-] = x and [HA] will become 0.02 - x. However, as Ka is very small, x will be very small compared to 0.02, thus we can approximate [HA] as 0.02 M. The equilibrium equation is Ka = (x)(x)/(0.02).
03

Solve for x

Plug in the value of Ka into the equilibrium equation to find x (the concentration of H+). Ka = (2 x 10^-12) = (x)(x)/(0.02). By solving this equation, x^2 = (2 x 10^-12)(0.02). Thus, x^2 = 4 x 10^-14 and x = 2 x 10^-7 M.
04

Calculate the pH

The pH of the solution is -log[H+]. Substitute x into this equation: pH = -log(2 x 10^-7).
05

Simplify the pH Calculation

By using logarithm properties: pH = -log(2) - log(10^-7) = -0.3 + 7 = 6.7.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acid Dissociation Constant
Understanding how a weak acid dissociates in water is crucial to calculating the pH of its solution. The acid dissociation constant, often represented as Ka, measures the strength of an acid in solution. It indicates the equilibrium between the undissociated acid (HA) and the ions it forms, commonly H+ and A-. For the equation HA ↔ H+ + A-, the Ka is expressed mathematically as:

\( K_a = \frac{[H^+][A^-]}{[HA]} \).

This equation implies that the higher the Ka, the stronger the acid because it dissociates more in water. Conversely, a smaller Ka value signifies a weaker acid. In our case, with a Ka of \(2 \times 10^{-12}\), HA can be classified as a weak acid, indicating it only partially dissociates in solution.
Equilibrium Constant
The equilibrium constant, in the context of acid-base reactions, is indeed the acid dissociation constant Ka. At equilibrium, the rate of the forward reaction (the dissociation of the weak acid) equals the rate of the reverse reaction (the recombination of the ions to form the weak acid). This condition is captured by the equilibrium expression.

Since the dissociation of a weak acid is a dynamic process, the concept of equilibrium allows us to calculate the exact concentrations of all species in solution at any given time, provided we know the initial concentration of the acid and the Ka value. It's important to note that in dilute solutions, where the acid dissociation is very small (Ka << 1), we often make the assumption that the initial concentration of the acid remains relatively unchanged, thus simplifying our calculations greatly.
pH of Solution
The pH of a solution quantifies its acidity or basicity on a logarithmic scale with values typically ranging from 0 to 14. A pH lower than 7 indicates an acidic solution, while a pH higher than 7 indicates a basic (or alkaline) solution, with 7 being neutral pH.

To calculate the pH of a weak acid solution, we first ascertain the concentration of hydrogen ions ([H+]) in the solution at equilibrium. As we saw in the exercise, once we have the H+ concentration, we apply the formula:

\( pH = -\log [H^+] \).

The 'p' in pH stands for 'potenz', which means power in German, and the H is for hydrogen. Therefore, pH is essentially telling us the 'power' or concentration of hydrogen ions present in a solution. The lower the concentration of H+ (i.e., the higher the pH), the less acidic the solution is, and vice versa.
Logarithm Properties
Essential to pH calculations are properties of logarithms, which are the mathematical tools used to transform the multiplication and division of numbers into addition and subtraction problems, easing the process of calculation. Logarithms have several properties that are commonly used:

  • The product rule: \(\log(a \cdot b) = \log(a) + \log(b)\)
  • The quotient rule: \(\log(\frac{a}{b}) = \log(a) - \log(b)\)
  • The power rule: \(\log(a^b) = b \cdot \log(a)\)
  • If log base 10 is used (as in pH calculations), \(\log(10) = 1\).

Using these properties simplifies the pH calculation process substantially. For instance, we utilized the fact that \(\log(10^{-7}) = -7\) and therefore were able to calculate the pH by subtracting this value from the logarithm of the concentration of H+ ions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An acid type indicator, HIn differs in colour from its conjugate base \(\left(\mathrm{In}^{-}\right)\). The human eye is sensitive to colour differences only when the ratio \([\operatorname{In}] /[\mathrm{HIn}]\) is greater than 10 or smaller than \(0.1\). What should be the minimum change in the \(\mathrm{pH}\) of the solution to observe a complete colour change \(\left(K_{\mathrm{a}}=1.0 \times 10^{-5}\right)\) ? (a) \(0.0\) (b) \(1.0\) (c) \(2.0\) (d) \(5.0\)

What is the \(\mathrm{pH}\) of \(4 \times 10^{-3} \mathrm{M}-\mathrm{Y}(\mathrm{OH})_{2}\) solution assuming the first dissociation to be \(100 \%\) and second dissociation to be \(50 \%\), where \(Y\) represents a metal cation? \((\log 2=0.3, \log 3=0.48)\) (a) \(11.78\) (b) \(11.22\) (c) \(2.22\) (d) \(2.78\)

The solubility of sparingly soluble salt \(\mathrm{A}_{3} \mathrm{~B}_{2}\) (molar mass \(=\) 'M' \(\mathrm{g} / \mathrm{mol}\) ) in water is 'x' g/L. The ratio of molar concentration of \(\mathrm{B}^{3}\) to the solubility product of the salt is (a) \(\frac{108 x^{5}}{M^{5}}\) (b) \(\frac{x^{4}}{108 M^{4}}\) (c) \(\frac{x^{4}}{54 M^{4}}\) (d) \(\frac{x^{3}}{27 M^{3}}\)

The \(\mathrm{pH}\) at the equivalence point when a solution of \(0.01 \mathrm{M}-\mathrm{CH}_{3} \mathrm{COOH}\) is titrated with a solution of \(0.01 \mathrm{M}-\mathrm{NaOH}\), is \(\left(\mathrm{p} K_{\mathrm{a}}\right.\) of \(\mathrm{CH}_{3} \mathrm{COOH}=4.7, \log 5=0.7\) ) (a) \(10.5\) (b) \(3.5\) (c) \(10.35\) (d) \(3.65\)

The buffer capacity \((\beta)\) for a weak acid (A) \(-\) conjugate base (B) buffer is defined as the number of moles of strong acid or base needed to change the \(\mathrm{pH}\) of \(1 \mathrm{~L}\) of solution by \(1 \mathrm{pH}\) unit, where \(\beta=\frac{2.303\left(C_{\mathrm{A}}+C_{\mathrm{B}}\right) K_{\mathrm{a}}\left[\mathrm{H}^{+}\right]}{\left(\left[\mathrm{H}^{+}\right]+K_{\mathrm{a}}\right)^{2}} .\) Under what condition will a buffer best resist a change in \(\mathrm{pH}\) ? (a) \(\mathrm{pH}=3 \mathrm{p} \mathrm{Ka}\) (b) \(2 \mathrm{pH}=\mathrm{p} \mathrm{Ka}\) (c) \(\mathrm{pH}=\mathrm{p} \mathrm{Ka}\) (d) \(\mathrm{pH}=2 \mathrm{p} \mathrm{Ka}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free