All categories

3 years ago

A 0.085 M solution of a weak acid has a pH of 2.68. Determine the Ka value for this acid.

Chemistry

1 answer:

lions [1.4K]3 years ago

4 0

HOPE THIS HELPS YOU!!!

You might be interested in

The concentration of hydroxide ions is less than the concentration of hydronium ions for acidic solutions. True False

wolverine [178]

Answer:

True

Explanation:

It's true because the pH is a measure of how basic or acid a solution is. In an acidic medium, the pH scales goes from 0 to 7. While in a basic medium goes from 7 to 14. The lower the pH value of the most acid the solution is.

1. The expression pH = -log(molar concentration of hydronium) allow to calculate the pH of a solution.

2. On the other hand, the expression pOH = -log(molar concentration of hydroxide) allow to determine the pOH of a solution.

The values of pH and pOH always obey the following expression:

pH + pOH = 14

Thus if for instance the pH becomes smaller the pOH must become bigger in order to fulfill the equation. Which means that the concentration of hydronium ions is greater than the hydroxide concentration.

For example, in an acidic medium:

if pH= 3, pOH= 11

In this case the molar concentration of hydronium is 0,001M. And the molar concentration of hydroxide ions is just 0,00000000001M.

4 0

3 years ago

Read 2 more answers

HELP FAST!!!!

Viktor [21]

Try d)kinetic energy, if wrong sorry

4 0

3 years ago

An overhead view of two people at the start of a maze. The person in a purple shirt has a line running forward, right, backward,

Lubov Fominskaja [6]

Answer:

Zamir walked

meters and Talia walked

meters.

Explanation:

27 and 19

8 0

3 years ago

Read 2 more answers

Calculate Delta H in KJ for the following reactions using heats of formation:

lozanna [386]

Answer:

$\Delta H\textdegree = -2856.8\;\text{kJ}$ per mole reaction.

$\Delta H\textdegree = -22.3\;\text{kJ}$ per mole reaction.

Explanation:

What is the standard enthalpy of formation $\Delta H_f\textdegree{}$ of a substance? $\Delta H_f\textdegree{}$ the enthalpy change when one mole of the substance is formed from the most stable allotrope of its elements under standard conditions.

Naturally, $\Delta H_f\textdegree{} = 0$ for the most stable allotrope of each element under standard conditions. For example, oxygen $\text{O}_2$ (not ozone $\text{O}_3$ ) is the most stable allotrope of oxygen. Also, under STP $\text{O}_2$ is a gas. Forming $\text{O}_2\;(g)$ from itself does not involve any chemical or physical change. As a result, $\Delta H_f\textdegree{} = 0$ for $\text{O}_2\;(g)$ .

Look up standard enthalpy of formation $\Delta H_f\textdegree{}$ data for the rest of the species. In case one or more values are not available from your school, here are the published ones. Note the state symbols of the compounds (water/steam $\text{H}_2\text{O}$ in particular) and the sign of the enthalpy changes.

$\text{C}_2\text{H}_6\;(g)$ : $-84.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{CO}_2\;(g)$ : $-393.5\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{H}_2\text{O}\;{\bf (g)}$ : $-241.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}\;(s)$ : $-217.9\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{PbO}_2\;(s)$ : $-276.6\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\text{Pb}_3\text{O}_4\;(s)$ : $-734.7\;\text{kJ}\cdot\text{mol}^{-1}$

How to calculate the enthalpy change of a reaction $\Delta H_\text{rxn}$ (or simply $\Delta H$ from enthalpies of formation?

Multiply the enthalpy of formation of each product by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Reactants}))$ to show that this value takes the coefficients into account.
Multiply the enthalpy of formation of each reactant by its coefficient in the equation.
Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Products}))$ to show that this value takes the coefficient into account.
Change = Final - Initial. So is the case with enthalpy changes. $\Delta H_\text{rxn} = \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))$ .

For the first reaction:

$\Sigma (n\cdot \Delta_f(\text{Reactants})) = 4\times (-393.5) + 6\times (-241.8) = -3024.8\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\Sigma (n\cdot \Delta_f(\text{Products})) = 2\times (-84.0) + 7\times 0 = -168.0\;\text{kJ}\cdot\text{mol}^{-1}$ ;
$\begin{aligned}\Delta H_\text{rxn} &= \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))\\ &= (-3024.8\;\text{kJ}\cdot\text{mol}^{-1}) - (-168.0\;\text{kJ}\cdot\text{mol}^{-1})\\ &= -2856.8\;\text{kJ}\cdot\text{mol}^{-1} \end{aligned}$ .