Hydrocarbons separated by fractional distillation of petroleum can be cracked to make

Do parts a, b and c

Leona [35]

Answer:- (a)The pH of the buffer solution is 3.90.

(b) the pH of the solution after addition of HCl would be 3.60.

(c) the pH of the buffer solution after addition of NaOH is 4.32.

Solution:- (a) It is a buffer solution so the pH could easily be calculated using Handerson equation:

$pH=Pka+log(\frac{base}{acid})$

pKa can be calculated from given Ka value as:

$pKa=-logKa$

$pKa=-log(6.3*10^-^5)$

pKa = 4.20

let's plug in the values in the Handerson equation:

$pH=4.20+log(\frac{0.025}{0.05})$

pH = 4.20 - 0.30

pH = 3.90

The pH of the buffer solution is 3.90.

(b) Let's say the acid is represented by HA and the base is represented by $A^-$ .

Original mili moles of HA from part a = 0.05(100) = 5

original mili moles of $A^-$ from part a = 0.025(100) = 2.5

mili moles of HCl that is $H^+$ added = 0.100(10.0) = 1

This HCl reacts with the base present in the buffer to make HA as:

$A^-+H^+\rightarrow HA$

Total mili moles of HA after addition of HCl = 5+1 = 6

mili moles of base after addition of HCl = 2.5-1 = 1.5

Let's plug in the values in the Handerson equation again. Here, we could use the mili moles to calculate the pH. The answer remains same even if we use the concentrations also as the final volume is same both for acid and base.

$pH=4.20+log(\frac{1.5}{6})$

pH = 4.20 - 0.60

pH = 3.60

So, the pH of the solution after addition of HCl would be 3.60.

(c) mili moles of NaOH or $OH^-$ added to the original buffer = 0.05(15.0) = 0.75

This $OH^-$ reacts with HA to form $A^-$ as:

$HA+OH^-\rightarrow H_2O+A^-$

mili moles of HA after addition of NaOH = 5-0.75 = 4.25

mili moles of $A^-$ after addition of NaOH = 2.5+0.75 = 3.25

Let's plug in the values again in Handerson equation:

$pH=4.20+log(\frac{3.25}{4.25})$

pH = 4.20 - 0.12

pH = 4.32

So, the pH of the buffer solution after addition of NaOH is 4.32.

7 0

3 years ago

Is a substances ability to conduct energy a physical or chemical property?

STatiana [176]

Chemical Property is the answer

6 0

3 years ago

Assume we have 25 mL of a 2.5 M HCl and a solution of NaOH of unkown concentration. The two react via the following equation.

navik [9.2K]

Answer is A - HCl is the titrant and D-NaOH is the analyte.

In a titration process, the solution of unknown concentration considered is the analyte. During titration, a standard solution(titrant) is added to an analyte until the equivalence point is achieved.

So in this case as the concentration of NaOH is not known,NaOH is the analyte.

Titrant is the solution whose concentration is known to us and is added to an analyte until the equivalence point is reached. In this case since we know the concentration of HCl, HCl is the titrant.

6 0

3 years ago

When 4.50 L of hydrogen gas react with an excess of nitrogen gas at standard temperature and pressure, how many liters of ammoni

ehidna [41]

<span>If the gasses participates in the reaction, and they under the same temperature and pressure, ratio of their volumes is equal to ratio of the moles of these gasses in the reaction.
N2 (g) + 3H2 (g)-------> 2NH3 (g)
from reaction 3 mol 2 mol
from the problem 4.5L x L

x=4.5*⅔=9/3=3 L<span>
</span></span>

8 0

3 years ago

How many valence electrons are represented in the following electron configuration? 1s22s22p5

dalvyx [7]

The element above has 7 (OPTION D) valence electrons. This is because on the second shell, two electrons occupy the S orbital of the second shell (2s²)and five occupy the P orbital (2p⁵) , for a total of 7 valence electrons (1s² 2s² 2p⁵)

4 0

3 years ago