All categories

3 years ago

Can you make a solution with a pH of 7 by changing ions?

1 answer:

6 0

No, because NaCl is a neutral salt and doesn't affect the pH. Obviously such a high salt concentration change the chemical activity of the ions so the acid-base equilibrium will change. Even if you had added a strong acid would you see the effect of salt as pH is defined as the log to the activity of H+.

You might be interested in

When a sulfur atom gains its valence electrons to have 8, the charge on the resulting ion is

nikklg [1K]

When a sulfur atom gains its valence electrons to have 8, the charge on the resulting ion is stable since sulfur atom achieved the octet rule. An octet rule is the filling of 8 electrons in an atom.

8 0

4 years ago

Read the statement.

lina2011 [118]

Answer:

60 g/L is the final concentration of NaI solution .

Explanation:

Molarity of NaI solution before evaporation = $M_1= 0.2 M$

Volume of NaI solution before evaporation = $V_1= 2.0 L$

Molarity of NaI solution after evaporation = $M_2= ?$

Volume of NaI solution after evaporation = $V_2= 1.0 L$

$M_1V_1=M_2V_2$ ( dilution)

$M_2=\frac{M_1V_1}{V_2}=\frac{0.2 M\times 2.0 L}{1.0 L}=0.4 M$

Molar mass of NaI = 150 g/mol

Concentration of NaI after evaporation :

0.4 M × 150 g/mol = 60 g/L

60 g/L is the final concentration of NaI solution .

4 0

3 years ago

Which of the following is a salt? A. O2 B. C2H2 C. C6H12O6 D. NaCl

Aleks04 [339]

O2 would be your answer.

3 0

4 years ago

Read 2 more answers

Ozone, o3, is a product in automobile exhaust by the reaction represented by the equation no2(g) + o2(g) --> 3no(g) + o3(g).

svetoff [14.1K]

Answer: 2.1 g mass of ozone( $O_{3}$ ) is predicted to form from the reaction of 2.0 g $NO_{2}$ in a car's exhaust and excess oxygen

Given information : Mass of $NO_{2}$ = 2.0 g and $O_{2}$ is in excess.

We need to calculate the mass of ozone ( $O_{3}$ )

Mass of ozone( $O_{3}$ ) is calculated with the help of mass of $NO_{2}$ using stoichiometry.

$NO_{2} + O_{2}\rightarrow NO + O_{3}$

Step 1 : Convert grams of $NO_{2}$ to moles of $NO_{2}$ .

$Moles = \frac{Grams}{Molar mass}$

Molar mass of $NO_{2}$ = 46.0 g/mol

$Moles = \frac{2.0g}{46.0\frac{g}{mol}}$

Moles of $NO_{2}$ = 0.043 mol

Step 2 : Find the moles of $O_{3}$ using moles of $NO_{2}$ .

Moles of $O_{3}$ is calculated by using moles of $NO_{2}$ with the help of mole ratio.

A mole ratio is the ratio between the amounts in moles of any two compounds involved in a chemical reaction. The mole ratio may be determined by examining the coefficients in front of formulas in a balanced chemical equation.

From the balanced chemical equation we can see that coefficient of $NO_{2}$ is 1 and coefficient of O3 is 1 , so mole ratio of $O_{3}$ to $NO_{2}$ is 1:1

Moles of $O_{3}$ = $(0.043 mol NO_{2})\times \frac{(1 mol O_{3})}{(1 mol NO_{2})}$

Moles of $O_{3}$ = $(0.043)\times \frac{(1 mol O_{3})}{(1)}$

Moles of $O_{3}$ = 0.043 mol

Step 3 : Convert moles of $O_{3}$ to grams of $O_{3}$

Grams = Moles X Molar mass

Molar mass of $O_{3}$ = 48.0 g/mol

Grams = $(0.043 mol O_{3})\times (\frac{48 g O_{3}}{1 mol O_{3}})$

Grams = $(0.043)\times (\frac{48 g O_{3}}{1})$

Grams = 2.1 g $O_{3}$

Note : The above three steps can also be done using a single step setup.

Grams of $O_{3}$ = $(2.0 gNO_{2})\times \frac{(1mol NO_{2})}{(46.0 g NO_{2})}\times \frac{(1 mol O_{3})}{(1 mol NO_{2})}\times \frac{(48.0 g O_{3})}{(1 mol O_{3})}$