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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

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What is the pH of a 3.5 x 10 to the negative 11th M H+ solution

Ghella [55]

Answer:

Explanation:

-log(3.5 * 10^-11)

= 10.4559

Be careful how you put this into your calculator. I had to use Exp to get it to work properly.

-log

(3.5 * 10 exp -11)

4 0

3 years ago

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Why do the planets keep orbiting the sun

dsp73

There's no reason, it's just because the sun has a gravitational pull

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3 years ago

PLEASE HELP ASAP!

choli [55]

Answer:

$\large \boxed{\text{D. 710 g}}$

Explanation:

1. Calculate the molar mass of Na₂SO₄

$\begin{array}{ccc}\textbf{Atoms} &\textbf{M}_{\textbf{r}} & \textbf{Mass/u}\\\text{2Na} & 23 & 46\\\text{1S} & 32 & 32\\\text{4O}&16 & 64\\&\text{TOTAL =} & \mathbf{142}\\\end{array}$

The molar mass of Na₂SO₄ is 142 g/mol.

2. Calculate the moles of Na₂SO₄

$\text{Moles of Na$_{2}$SO}_{4} = \text{2.5 L solution} \times \dfrac{\text{2.0 mol Na$_{2}$SO}_{4}}{\text{1 L solution}} = \text{5.0 mol Na$_{2}$SO}_{4}$

3. Calculate the mass of Na₂SO₄

$\text{Mass of Na$_{2}$SO}_{4} = \text{5.0 mol Na$_{2}$SO}_{4} \times \dfrac{\text{142 g Na$_{2}$SO}_{4}}{\text{1 mol Na$_{2}$SO}_{4}} = \text{710 g Na$_{2}$SO}_{4}\\\\\text{You need } \large \boxed{\textbf{710 g}} \text{ of Na$_{2}$SO}_{4}$

6 0

2 years ago

NaOH + X mc005-1.jpg NaCH3COO + H2O What is X in this reaction?

Allushta [10]

"X" in the reaction above is acetic acid with a chemical formula CH3COOH. The chemical reaction would be NaOH + CH3COOH = NaCH3COO + H2O. This is a neutralization reaction in which it produces a salt and water. The salt produced is called sodium acetate.

7 0

2 years ago

The mass of sodium chloride in (g) is 14.19 The volume of ammonia solution in (mL) is 36.15 Calculate the following: What is the

Svetach [21]

This is an incomplete question, here is a complete question.

A weighed amount of sodium chloride is completely dissolved in a measured volume of 4.00 M ammonia solution at ice temperature, and carbon dioxide is bubbled in. Assume that sodium bicarbonate is formed util the limiting reagent is entirely used up. the solubility of sodium bicarbonate in water at ice temperature is 0.75 mol per liter. Also assume that all the sodium bicarbonate precipitated is collected and converted quantitatively to sodium carbonate.

Data to be used for calculating the results

-The mass of sodium chloride in (g) is 14.19

-The volume of ammonia solution in (mL) is 36.15

Calculate the following: What is the theoretical yield of sodium bicarbonate in grams?

Answer : The theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

Explanation :

First we have to calculate the moles of NaCl and $NH_3$ .

$\text{ Moles of }NaCl=\frac{\text{ Mass of }NaCl}{\text{ Molar mass of }NaCl}=\frac{14.19g}{58.5g/mole}=0.243moles$

$\text{ Moles of }NH_3=\text{ Concentration of }NH_3\times \text{ Volume of solution}=4.00M\times 0.3615L=1.446moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$NH_3+NaCl+CO_2+H_2O\rightarrow NaHCO_3+NH_4Cl$

From the balanced reaction we conclude that

As, 1 mole of $NaCl$ react with 1 mole of $NH_3$

So, 0.243 mole of $NaCl$ react with 0.243 mole of $NH_3$

From this we conclude that, $NH_3$ is an excess reagent because the given moles are greater than the required moles and $NaCl$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaHCO_3$

From the reaction, we conclude that

As, 1 mole of $NaCl$ react to give 1 mole of $NaHCO_3$

So, 0.243 moles of $NaCl$ react to give 0.243 moles of $NaHCO_3$

Now we have to calculate the mass of $NaHCO_3$

$\text{ Mass of }NaHCO_3=\text{ Moles of }NaHCO_3\times \text{ Molar mass of }NaHCO_3$

Molar mass of sodium bicarbonate = 84 g/mol

$\text{ Mass of }NaHCO_3=(0.243moles)\times (84g/mole)=20.4g$

Thus, the theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

4 0

3 years ago