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3 years ago

What is the hydroxide concentration of a solution with a pH of 12.80?

1 answer:

8 0

Step1 : To convert from pH to ion concentrations,first apply equation 17-1 to calculate [H3O+]. Then make use of water equilibrium to calculate [OH-]
Step 2 : We must rearrange equation pH = -log [H+] , in order to solve for concentration
Step 3 : log [H+] = -pH and [H+] = 10 -pH
Step 4 ; Now substitute and evaluate:
[H+] = 10-3.05 = 8.9 x 10-4 M
Step 5: To determine the hydroxide ion concentration,we need to rearrange the water equilibrium expression:
Kw = 1.0 x 10 -14 = [H3O+]eq [OH-] and [OH-]eq = 1.0 x 10- 14 / [H3O+]
now susbtitute and evaluate:
= 1.0 x 10- 14 / 8.9 x 10-4 = 1.1 x 10-11 M

Sample protblem 2 : What is the OH- ion concetraion of a solution of NaOH with pH 13.40 ?
Solution:
Step 1 : First find pOH from the given pH:
pH + pOH = 14
pOH = 14 - 13.40 = 0.60
Step 2 : As you have now the value of pOH you can find the OH- ion concentration using formula
pOH = -log [OH-]
[OH- ] = 10 - pOH = 10 - 0.60 = 0.25 M

You might be interested in

If I need 2.2 moles of CO2 , and I have excess Fe2O3 , how many moles of C do I need?

olchik [2.2K]

Answer:

0.733 mol.

Explanation:

From the balanced equation:

2Fe₂O₃ + C → Fe + 3CO₂,

It is clear that 1.0 moles of Fe₂O₃ react with 1.0 mole of C to produce 1.0 mole of Fe and 3.0 moles of CO₂.

Since Fe₂O₃ is in excess, C will be the limiting reactant.

Using cross multiplication:

1.0 mole of C produces → 3.0 moles of CO₂, from the stichiometry.

??? mole of C produces → 2.2 moles of CO₂.

∴ The no. of moles of C needed to produce 2.2 moles of CO₂ = (1.0 mole of C) (2.2 mole of CO₂) / (3.0 mole of CO₂) = 0.733 mol.

6 0

2 years ago

29. A gas has a volume of 1.75 L at -23°C and 150.0 kPa.

arsen [322]

The answer for the following mention bellow.

Therefore the final temperature of the gas is 260 k

Explanation:

Given:

Initial pressure ( $P_{1}$ ) = 150.0 kPa

Final pressure ( $P_{2}$ ) = 210.0 kPa

Initial volume ( $V_{1}$ ) = 1.75 L

Final volume ( $V_{2}$ ) = 1.30 L

Initial temperature ( $T_{1}$ ) = -23°C = 250 k

To find:

Final temperature ( $T_{2}$ )

We know;

According to the ideal gas equation;

P × V = n × R ×T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

We know;

$\frac{P*V}{T}$ = constant

$\frac{P_{1} }{P_{2} }$ × $\frac{V_{1} }{V_{2} }$ = $\frac{T_{1} }{T_{2} }$

Where;

( $P_{1}$ ) represents the initial pressure of the gas

( $P_{2}$ ) represents the final pressure of the gas

( $V_{1}$ ) represents the initial volume of the gas

( $V_{2}$ ) represents the final volume of the gas

( $T_{1}$ ) represents the initial temperature of the gas

( $T_{2}$ ) represents the final temperature of the gas

So;

$\frac{150 * 1.75}{210 * 1.30}$ = $\frac{260}{T_{2} }$

( $T_{2}$ ) =260 k

Therefore the final temperature of the gas is 260 k

3 0

2 years ago

Please help Please help

alexdok [17]

Follow Avogadro’s Number
1 mole = 6.02 x 10^23
So we can do it
4.77x10^25/6.02x10^23 = 79.2 mole

5 0

1 year ago

Which of the following represents the smallest mass? I.) 23 cg II.) 2.3 x 10-3 μg III.) 0.23 mg IV.) 0.23 g V.) 2.3 x 10-2 kg

jenyasd209 [6]

The smallest mass among the choices is Choice (ii):2.3 x 10-3 μg

This is so because, the prefix μ signifies a factor of 10^-6.

For choice I:

23 cg means 23 centigrams

In essence = 0.23grams

For choice II:

2.3 × 10-3 μg means 2.3 × 10-9 grams

For choice III:

0.23 mg means 0.23 × 10-3 grams

For choice IV:

0.23 grams simply means 0.23grams

For choice V:

2.3 × 10-2 kg means 23 grams

Ultimately, Choice II represents the smallest mass.