What is the pH of a 0.600 M solution of HSo, (Ka = 6.6 × 10°)?(Please show your answers with 3 sig figs.)| M1346.C789+/-x 100

Answered: What is the pH of a 0.600 M solution of…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Concept explainers

Question

What is the pH of a 0.600 M solution of HSo, (Ka = 6.6 × 10°)?
(Please show your answers with 3 sig figs.)
| M
1
3
4
6.
C
7
8
9
+/-
x 100

Expert Solution

Step 1

pH of solution is equal to the negative logarithm of hydronium ion concentration.

$pH = - \log (H_{3} O^{+})$

K_a is known as acid dissociation constant. It is equal to the ratio of concentration of product to the concentration of reactant.

Higher the value of K_a, higher the acidic nature.

Significant numbers means meaningful digits.

Step 2

To determine pH of given solution of HSO₃^-.

Data given.

Concentration of HSO₃^-=0.600M.

Acid dissociation reaction of HSO₃^-.

${HSO}_{3}^{-} (a q) \to_{}^{} H^{+} (aq) + {SO}_{4}^{2 -} (a q) K_{a} = \frac{[{SO}_{4}^{2 -}] [H^{+}]}{[{HSO}_{3}^{-}]}$

Let x be the concentration of sulfate and H⁺ ions at equilibrium.

${HSO}_{3}^{-} (aq) \to_{}^{} H^{+} (aq) + {SO}_{3}^{2 -} (aq) I 0.600 M 0 0 C - x x x E 0.600 M - x x x$

$K_{a} = \frac{{(x)}^{2}}{(0.600 - x)} 6.6 \times 10^{- 8} = \frac{{(x)}^{2}}{(0.600 - x)} x^{2} + 6.6 \times 10^{- 8} x - 3.96 \times 10^{- 8} = 0 .$

on solving quadratic equation.

x=0.0001989M.

Concentration of H⁺ ion is 0.0001989M.

$pH = - \log (H^{+}) . pH = - \log (0.0001989) . pH = 3.70$

Step by step

Solved in 3 steps

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions