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Calculate the pH of a solution in which one normal adult dose of aspirin (640 mg ) is dissolved in 10 ounces of water. Express y

d1i1m1o1n [39]

The pH of the solution in which one normal adult dose aspirin is dissolved is : 2.7

Given data :

mass of aspirin = 640 mg = 0.640 g

volume of water = 10 ounces = 0.295735 L

molar mass of aspirin = 180.16 g/mol

moles of aspirin = mass / molar mass = 0.00355 mol

<h3>Determine the pH of the solution </h3>

First step : <u>calculate the concentration of aspirin</u>

= moles of Aspirin / volume of water

= 0.00355 / 0.295735

= 0.012 M

Given that pKa of Aspirin = 3.5

pKa = -logKa

therefore ; Ka = $10^{-3.5}$ = $3.162 * 10^{-4}$

From the Ice table

$3.162 * 10^{-4}$ = $\frac{x + H^+}{[aspirin]}$ = $\frac{x^{2} }{0.012-x}$

given that the value of Ka is small we will ignore -x

x² = $3.162 * 10^{-4} * 0.012$

x = $1.948 * 10^{-3}$

Therefore

[ H⁺ ] = $1.948 * 10^{-3}$

given that

pH = - Log [ H⁺ ]

= - ( -3 + log 1.948 )

= 2.71 ≈ 2.7

Hence we can conclude that The pH of the solution in which one normal adult dose aspirin is dissolved is : 2.7

Learn more about Aspirin : brainly.com/question/2070753

4 0

2 years ago

BRAINLIESTTT ASAP!!! PLEASE HELP ME :)

lapo4ka [179]

Steps/Answers:

Jot down the metal name.

Write the metal name ending in -idle.

Combine cation, annion names,

Hope This Helps! Have A Nice Day!!

5 0

3 years ago

Give an example of an element that would be expected to share similar chemical properties with beryllium, calcium, and strontium

Aleks04 [339]

Strontium possibly but I’m not 100% sure

4 0

3 years ago

28) Consider a 21.0 mL sample of pure lemon juice with a citric acid (H3C6H5O7) concentration of 0.30M. a. How many moles of cir

Damm [24]

<h3>#a. Answer:</h3>

0.0063 mole

<h3>Solution and explanation:</h3>

We are given 21.0 mL citric acid with a concentration of 0.30 M

Part a requires we calculate the number of moles of citric acid.

We need to know how to calculate the concentration of a solution;

Concentration or molarity = Number of moles ÷ Volume of the solution

Thus;

Number of moles = Concentration × Volume

Hence;

Moles = 0.30 M × 0.021 L

= 0.0063 mole

<h3>#b. Answer</h3>

1.21 g citric acid

<h3>Solution</h3>

Part B

We are required to calculate the mass of citric acid in the sample

Number of moles of a compound is calculated by dividing its mass by its molar mass.

Molar mass of Citric acid = 192.124 g/mol

Moles of citric acid = 0.0063 mole

But; Mass = Number of moles × Molar mass

Mass of citric acid = 0.0063 mol × 192.124 g/mol

= 1.21 g citric acid

<h3>#c. Answer</h3>

4.167 mL

<h3>Solution:</h3>

Part C

We are required to determine the initial volume before dilution;

We have;

Initial concentration (M1) = 0.30 M

Final volume (V2) = 250 mL or 0.25 L

Final concentration (M2) = 0.0050 M

Using the dilution formula we can get the initial volume;

Therefore, since; M1V1 =M2V2

V1 = M2V2÷M1

= (0.0050 × 0.25)÷ 0.30

= 0.004167 L or

= 4.167 mL

Therefore, the initial volume of the solution is 4.167 mL

8 0

3 years ago

Phosphorous trichloride and phosphorous pentachloride equilibrate in the presence of molecular chlorine according to the reactio

umka21 [38]

Answer : The value of $K_p$ at this temperature is 66.7

Explanation : Given,

Pressure of $PCl_3$ at equilibrium = 0.348 atm

Pressure of $Cl_2$ at equilibrium = 0.441 atm

Pressure of $PCl_5$ at equilibrium = 10.24 atm

The balanced equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The expression of equilibrium constant $K_p$ for the reaction will be:

$K_p=\frac{(p_{PCl_5})}{(p_{PCl_3})(p_{Cl_2})}$

Now put all the values in this expression, we get :

$K_p=\frac{(10.24)}{(0.348)(0.441)}$

$K_p=66.7$

Therefore, the value of $K_p$ at this temperature is 66.7

4 0

3 years ago

Bantu donggg Aku butuh banget Makasih.....