Ask question

Ask question

All categories

3 years ago

5

Suppose that you have 135 mL of a buffer that is 0.360 M in both propanoic acid ( C 2 H 5 COOH ) and its conjugate base ( C 2 H

5 COO − ) . Calculate the maximum volume of 0.240 M HCl that can be added to the buffer before its buffering capacity is lost.

1 answer:

Minchanka [31]3 years ago

5 0

202.50 ml is the maximum volume of 0.240 M HCl that can be added to the buffer before its buffering capacity is lost.

Explanation:

Data given:

volume of buffer = 135 ml or 0.135 litres

molarity of the buffer = 0.360 M

volume of the acid = ?

molarity of the acid = 0.24 M

the number of moles of conjugate base is

M X V = number of moles

0.360 X 0.135

0.0486 moles of conjugate base

The maximum amount of acid added will equal the amount of conjugate base from the buffer.

So, the volume of acid will be calculated by using the formula:

Molarity = $\frac{number of moles}{volume}$

volume = $\frac{0.0486}{0.24}$

= 0.2025 l

202.50 ml is the volume of acid added.

You might be interested in

According to Neil’s Bohr’s atomic model , what occurs when an atom absorbs radiated energy?

sergiy2304 [10]

An electron shifts from a lower energy to a higher energy

3 0

3 years ago

AP chem. Sublevels!!!!!!!!!!!

slavikrds [6]

Answer:

2d

Explanation:

The lowest d is 3d.

Hope this helps!

5 0

2 years ago

A steel cylinder has a pressure of 550 torr at 177 degrees Celsius. What will be the pressure if the temperature is decreased by

Schach [20]

Answer:

$P_2=507.7torr$

Explanation:

Hello there!

In this case, when we have a gas that is undergoing a change in both pressure and temperature, we utilize the Gay-Lussac's equation as shown below:

$\frac{T_2}{P_2}=\frac{T_1}{P_1}$

Thus, since we are given the initial pressure and temperature and the final temperature, we can compute the final pressure as shown below:

$P_2=\frac{P_1}{T_1}*T_2$

So we plug in, by making sure the temperatures are in kelvins, to obtain:

$P_2=\frac{550torr}{(117+273)K}*(87+273)K\\\\P_2=507.7torr$

Best regards!

8 0

3 years ago

A small 19 kilogram canoe is floating downriver at a speed of 5 m/s. What is the canoe's kinetic energy?

Oksanka [162]

the kinetic energy is the same as the canoes weight times its speed times the force of gravity.

4 0

3 years ago

Consider the reaction to produce methanolCO(g) + 2H2 (g) <-----> CH3OHAn equilibrium mixture in a 2.00-L vessel is found t

MariettaO [177]

Answer : The value of $K_c$ of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of $CH_3OH$ at equilibrium = 0.0406 mole

Moles of $CO$ at equilibrium = 0.170 mole

Moles of $H_2$ at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of $CH_3OH,CO\text{ and }H_2$ at equilibrium.

$\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M$

$\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M$

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M$

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

$CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)$

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

$K_c=\frac{[CH_3OH]}{[CO][H_2]^2}$

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

$K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}$

$K_c=10.5$

Therefore, the value of $K_c$ of the reaction is, 10.5

There are 3 conditions:

When $K_{c}>1$ ; the reaction is product favored.

When $K_{c}$ ; the reaction is reactant favored.

When $K_{c}=1$ ; the reaction is in equilibrium.

As the value of $K_{c}>1$ . So, the reaction is product favored.

7 0

3 years ago