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Elza [17]
3 years ago
5

What is the pH if 1mL of 0.1M HCl is added to 99mL of pure water?

Chemistry
1 answer:
coldgirl [10]3 years ago
5 0

Answer:

pH of buffer after addition of 1 mL of 0,1 M HCl = 7,0

Explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

pH = pka + log₁₀

Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺    Kₐ₂ = 6,20x10⁻⁸; pka=7,2

Thus, Henderson–Hasselbalch equation for 7,00 phosphate buffer is:

7,0 = 7,2 + log₁₀ \frac{[HPO4^{2-}] }{[H2PO4^{-}]}

Ratio obtained is:

0,63 = \frac{[HPO4^{2-}] }{[H2PO4^{-}]}

As the problem said you can assume [H₂PO₄⁻] = 0,1 M and [HPO4²⁻] = 0,063M

As the amount added of HCl is 0,001 M the concentrations in equilibrium are:

H₂PO₄⁻   ⇄   HPO4²⁻ +        H⁺

0,1 M +x      0,063M -x  0,001M -x -<em>because the addition of H⁺ displaces the equilibrium to the left-</em>

Knowing the equation of equilibrium is:

K_{a} = \frac{[HPO_{4}^{2-}][H^{+}]}{[H_{2} PO_{4}^{-}]}

Replacing:

6,20x10⁻⁸ = \frac{[0,063-x][0,001-x]}{[0,1+x]}

You will obtain:

x² -0,064 x + 6,29938x10⁻⁵ = 0

Thus:

x = 0,063 → No physical sense

x = 0,00099990

Thus, [H⁺] in equilibrium is:

0,001 M - 0,00099990 = 1x10⁻⁷

Thus, pH of buffer after addition of 1 mL of 0,1 M HCl =

-log₁₀ [1x10⁻⁷] = 7,0

A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. In this example you can see its effect!

I hope it helps!

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Answer:

V₂ = 12.43 L

Explanation:

Given data:

Initial pressure = 650 KPa

Initial volume = 2.2 L

Final pressure = 115 KPa

Final volume = ?

Solution:

The given problem will be solved through the Boyles law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

650 KPa ×2.2 L = 115 KPa × V₂

V₂ = 1430 KPa. L/ 115 KPa

V₂ = 12.43 L

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25cm3 of 0.2mol/dm3 barium hydroxide solution reacted with 22.8cm3 hydrochloric acid. Calculate the concentration of the hydroch
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Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide , carbon dioxide , n
mylen [45]

The question is incomplete, here is the complete question:

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide (CaO), carbon dioxide (CO_2), nitrogen dioxide (NO_2), and no other substances. A small sample gives 2.389 g CaO, 1.876 g CO_2, and 3.921 g NO_2 Determine the empirical formula of the compound.

<u>Answer:</u> The empirical formula for the given compound is CaCN_2

<u>Explanation:</u>

The chemical equation for the combustion of compound having calcium, carbon and nitrogen follows:

Ca_xC_yN_z+O_2\rightarrow CaO+CO_2+NO_2

where, 'x', 'y' and 'z' are the subscripts of calcium, carbon and nitrogen respectively.

We are given:

Mass of CaO = 2.389 g

Mass of CO_2=1.876g

Mass of NO_2=3.921g

We know that:

Molar mass of calcium oxide = 56 g/mol

Molar mass of carbon dioxide = 44 g/mol

Molar mass of nitrogen dioxide = 46 g/mol

<u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.876 g of carbon dioxide, \frac{12}{44}\times 1.876=0.5116g of carbon will be contained.

<u>For calculating the mass of nitrogen:</u>

In 46 g of nitrogen dioxide, 14 g of nitrogen is contained.

So, in 3.921 g of nitrogen dioxide, \frac{14}{46}\times 3.921=1.193g of nitrogen will be contained.

<u>For calculating the mass of calcium:</u>

In 56 g of calcium oxide, 40 g of calcium is contained.

So, in 2.389 g of calcium oxide, \frac{40}{56}\times 2.389=1.706g of calcium will be contained.

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Calcium =\frac{\text{Given mass of Calcium}}{\text{Molar mass of Calcium}}=\frac{1.706g}{40g/mole}=0.0426moles

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.5116g}{12g/mole}=0.0426moles

Moles of Nitrogen = \frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{1.193g}{14g/mole}=0.0852moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0426 moles.

For Calcium = \frac{0.0426}{0.0426}=1

For Carbon = \frac{0.0426}{0.0426}=1

For Nitrogen = \frac{0.0852}{0.0426}=2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of Ca : C : N = 1 : 1 : 2

Hence, the empirical formula for the given compound is CaCN_2

3 0
3 years ago
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