All categories

2 years ago

What is the pH of a solution that has a hydrogen ion concentration of 6.9*10^-4?

Chemistry

1 answer:

GarryVolchara [31]2 years ago

4 0

The pH would be 3.16 because the formula for finding the pH from a H± ion is pH = -log [H±]

So with that formula, you first have to substitute the H± with your numbers which in this case is 6.9 × 10^-4.

It should like this:

pH = -log (6.9 × 10^-4)

And lastly, just calculate.

pH = 3.16

Hope that helps, do let me know if you need more help in understanding this!

You might be interested in

What is the molarity of 53.0 grams of HCl dissolved in 400.0 mL of water

Bond [772]

Answer:

Molarity = 3.625 M

Explanation:

Given data:

Mass of HCl = 53.0 g

Volume of water = 400.0 mL (400/1000 = 0.4 L)

Molarity of solution = ?

Solution:

Number of moles of of HCl:

Number of moles = mass/ molar mass

Number of moles = 53 g/ 36.5 g/mol

Number of moles =1.45 mol

Molarity:

Molarity = number of moles of solute / volume of solution in L

Molarity = 1.45 mol / 0.4L

Molarity = 3.625 M

8 0

2 years ago

A student does an experiment to determine the molar solubility of lead(II) bromide. She constructs a voltaic cell at 298 K consi

attashe74 [19]

Answer:

The molar solubility of lead bromide at 298K is 0.010 mol/L.

Explanation:

In order to solve this problem, we need to use the Nernst Equaiton:

$E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}$

E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.

At equilibrium, E = 0, therefore:

$E^{o} = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] = log[ox] - \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] - \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 - \frac{2x5.45x10^{-2} }{0.0591}}\\\\$

[red] = 0.010 M

The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.

7 0

2 years ago

Consider the following intermediate chemical equations. When you form the final chemical equation, what should you do with CO? C

-Dominant- [34]

Answer:

Cancel out CO because it appears as a reactant in one intermediate reaction and a product in the other intermediate reaction.

Explanation:

The CO appears twice hence in he intermediate reaction it only forms path of the enabling reagents and it further reacts to form the final product. Accounting for the CO in the intermediate reaction that undergoes further reaction will impact on the stoichiometry of the reaction.

5 0

2 years ago

What were the names of the minerals and types of rocks?

padilas [110]

Types of rocks: Igneous, sedimentary, and metamorphic
There are over 200 names of minerals, I'm not sure what you want for that

8 0

2 years ago

Read 2 more answers

in the following reaction, how many grams of benzene (C6H6) will produce 42 grams of CO2? 2C6H6 + 15O2 → 12CO2 + 6H2O

Mrrafil [7]

Answer: -

12.41 g

Explanation: -

Mass of CO₂ = 42 g

Molar mass of CO₂ = 12 x 1 + 16 x 2 = 44 g / mol

Number of moles of CO₂ = $\frac{42}{44 g/mol}$

= 0.9545 mol

The balanced chemical equation for this process is

2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O

From the balanced chemical equation we see

12 mol of CO₂ is produced from 2 mol of C₆H₆

0.9545 mol of CO₂ is produced from $\frac{2 mol C6H6 x 0.9545 mol CO2}{12 mol CO2}$

= 0.159 mol of C₆H₆

Molar mass of C₆H₆ = 12 x 6 + 1 x 6 =78 g /mol

Mass of C₆H₆ =Molar mass x Number of moles

= 78 g / mol x 0.159 mol

= 12.41 g

8 0

2 years ago