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3 years ago

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If 20.0 ml of glacial acetic acid (pure hc2h3o2) is diluted to 1.40 l with water, what is the ph of the resulting solution? the

density of glacial acetic acid is 1.05 g/ml.

1 answer:

Y_Kistochka [10]3 years ago

8 0

By using the formula, mass = density x volume, we calculate mass in grams

20.0 mL CH₃COOH x (1.05 g / mL) = 21.0 g CH₃COOH

To find the moles, molar mass of CH₃COOH = 60.05g/mol<span>

21.0 g </span>CH₃COOH x (1 mole CH₃COOH / 60.05 g CH₃COOH) = 0.350 moles CH₃COOH

To find molarity,<span>

[</span>CH₃COOH] = moles CH₃COOH / L of solution = 0.350 / 1.40 = 0.250 M CH₃COOH<span>

When </span>CH₃COOH is dissolved in water, it produces small and equal amounts of H₃O⁺+ and C₂H₃O₂⁻.

<span>
Molarity , </span>CH₃COOH<span> + H</span>₂O <==> H₃O⁺ + C₂H₃O₂⁻

<span> <span>Initial 0.250 0 0 </span>
Change -x x x
Equilibrium 0.250-x x x

K</span>ₐ = [H₃O⁺][C₂H₃O₂⁻] / [HC₂H₃O₂] = (x)(x) / (0.250-x) = 1.8 x 10⁻⁵

<span>Since K</span>ₐ is relatively small, we can neglect the -x term after 0.250 to simplify

<span>x</span>² / 0.250 = 1.8 x 10⁻⁵

x² = 4.5 x 10⁻⁶

<span> x = 2.1 x 10</span>⁻³<span> = [H</span>₃O⁺]

pH = -log [H₃O⁺] = -log (2.1 x 10⁻³) = 2.68

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What is the molarity of a solution that contains 224 grams of KOH in 2<br> liters of solution?

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Answer:

$\boxed {\boxed {\sf 2 \ M \ KOH}}$

Explanation:

Molarity is a measure of concentration in moles per liter.

<h3>1. Grams to Moles </h3>

The first step is to convert the amount of grams given to moles. The molar mass is used. This found on the Periodic Table and it's the same value as the atomic mass, but the units are grams per mole.

We have 224 grams of KOH. Look up the molar masses for the individual elements.

Potassium (K): 39.098 g/mol
Oxygen (O): 15.999 g/mol
Hydrogen (H): 1.008 g/mol

Since the compound's formula has no subscripts, 1 formula unit has 1 atom of each element. We can simply add the molar masses together to find KOH's molar mass.

KOH: 39.098 + 15.999 + 1.008=56.105 g/mol

Use this number as a ratio.

$\frac {56.105 \ g\ KOH}{1 \ mol \ KOH}$

Multiply by the value we are converting: 224 g KOH

$224 \ g \ KOH *\frac {56.105 \ g\ KOH}{1 \ mol \ KOH}$

Flip the ratio so the units of grams KOH cancel.

$224 \ g \ KOH *\frac {1 \ mol \ KOH}{56.105 \ g\ KOH}$

$224 *\frac {1 \ mol \ KOH}{56.105}$

$\frac {224}{56.105} \ mol \ KOH$

$3.992514036 \ mol \ KOH$

<h3>2. Calculate Molarity </h3>

Remember molarity is moles per liter.

$molarity = \frac{moles}{liters}$

We just calculated the moles and we know there are 2 liters of solution.

$molarity = \frac{ 3.992514036 \ mol \ KOH}{ 2 \ L}$

$molarity= 1.996257018 \ mol \ KOH/ L$

<h3>3. Round and Convert Units </h3>

First, let's round. The original values have 3 and 1 significant figures. We go with the lowest number: 1. For the number we found, that is the ones place.

1.<u>9</u>96257018

The 9 in the tenths place tells us to round to 1 up to a 2

$2 \ mol \ KOH/ L$

Next, convert units. 1 mole per liter is equal to 1 molar or M.

$2 \ M \ KOH$

The molarity of the solution is <u>2 M KOH</u>

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Read 2 more answers

Use the given relationship to calculate the ionization energy of a thallium atom, given that radiation of 58.4 nm produces elect

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The ionization energy is the energy required to remove an electron from an atom.

We have the following information;

wavelength of the photon = 58.4 nm

Speed of the electron = 2310 × 10^3 m/s

Since;

hv = I + 1/2mv^2

v = c/λ

hc/λ = I + 1/2mv^2

I = hc/λ - 1/2mv^2

I = (6.6 × 10^-34 × 3 × 10^8/58.4 × 10^-9) - (1/2 × 9.11 × 10^-31 × (2310 × 10^3)^2)

I = 1 × 10^-18 J

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The mass of excess mass of $Na_2CO_3$ , $AgNO_3,Ag_2CO_3\text{ and }NaNO_3$ are, 1.908 g, 0 g, 12.144 g and 3.74 g respectively.

Explanation : Given,

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Mass of $AgNO_3$ = 7.50 g

Molar mass of $Na_2CO_3$ = 106 g/mole

Molar mass of $AgNO_3$ = 170 g/mole

Molar mass of $Ag_2CO_3$ = 276 g/mole

Molar mass of $NaNO_3$ = 85 g/mole

First we have to calculate the moles of $Na_2CO_3$ and $AgNO_3$ .

$\text{Moles of }Na_2CO_3=\frac{\text{Mass of }Na_2CO_3}{\text{Molar mass of }Na_2CO_3}=\frac{4.25g}{106g/mole}=0.040moles$

$\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=\frac{7.50g}{170g/mole}=0.044moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Na_2CO_3+2AgNO_3\rightarrow Ag_2CO_3+2NaNO_3$

From the balanced reaction we conclude that

As, 2 moles of $AgNO_3$ react with 1 mole of $Na_2CO_3$

So, 0.044 moles of $AgNO_3$ react with $\frac{0.044}{2}=0.022$ moles of $Na_2CO_3$

From this we conclude that, $Na_2CO_3$ is an excess reagent because the given moles are greater than the required moles and $AgNO_3$ is a limiting reagent and it limits the formation of product.

The excess mole of $Na_2CO_3$ = 0.040 - 0.022 = 0.018 mole

Now we have to calculate the mass of excess mole of $Na_2CO_3$ .

$\text{Mass of }Na_2CO_3=\text{Moles of }Na_2CO_3\times \text{Molar mass of }Na_2CO_3=(0.018mole)\times (106g/mole)=1.908g$

Now we have to calculate the moles of $Ag_2CO_3$ .

As, 1 moles of $AgNO_3$ react to give 1 moles of $Ag_2CO_3$

So, 0.044 moles of $AgNO_3$ react to give 0.044 moles of $Ag_2CO_3$

Now we have to calculate the mass of $AgCO_3$ .

$\text{Mass of }Ag_2CO_3=\text{Moles of }Ag_2CO_3\times \text{Molar mass of }Ag_2CO_3=(0.044mole)\times (276g/mole)=12.144g$

Now we have to calculate the moles of $NaNO_3$ .

As, 2 moles of $AgNO_3$ react to give 2 moles of $NaNO_3$

So, 0.044 moles of $AgNO_3$ react to give 0.044 moles of $NaNO_3$

Now we have to calculate the mass of $NaNO_3$ .

$\text{Mass of }NaNO_3=\text{Moles of }NaNO_3\times \text{Molar mass of }NaNO_3=(0.044mole)\times (85g/mole)=3.74g$

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4 years ago