Ask question

Ask question

All categories

Salsk061 [2.6K]

3 years ago

11

If a buffer is composed of 23.34 mL of 0.147 M acetic acid and 33.66 mL of 0.185 M sodium acetate, how many mL of 0.100 M NaOH c

an be added before the buffer capacity is reached

1 answer:

Leni [432]3 years ago

5 0

Answer:

25.5mL of 0.100M NaOH are needed to reach buffer capacity.

Explanation:

The buffer capacity is reached when the ratio between moles of conjugate base (Sodium acetate) and moles of weak acid (Acetic acid) is 10:

Moles sodium acetate / Moles Acetic acid = 10

The reaction of acetic acid, HA, with NaOH, to produce sodium acetate, NaA is:

HA + NaOH → H2O + NaA

That means the moles of NaOH added = Moles of HA that are being subtracted and moles of NaA that are been produced.

The initial moles of each species is:

<em>Acetic acid:</em>

23.34mL = 0.02334L * (0.147mol / L) = 0.00343 moles Acetic Acid

<em>Sodium Acetate:</em>

33.66mL = 0.03366L * (0.185mol / L) = 0.00623 moles Sodium Acetate

We can write the moles of each species when NaOH is added as:

Moles sodium acetate / Moles Acetic acid = 10

0.00623 moles + X / 0.00343 moles - X = 10

<em>Where X are moles of NaOH added</em>

Solving for X:

0.00623 moles + X = 0.0343 moles - 10X

11X = 0.0281

X = 0.00255 moles of NaOH are needed

In Liters:

0.0255mol NaOH * (1L / 0.100mol) = 0.0255L of 0.100M NaOH are needed =

<h3>25.5mL of 0.100M NaOH are needed to reach buffer capacity</h3>

<em />

You might be interested in

How many grams of sodium acetate ( molar mass = 83.06 g/mol ) must be added to 1.00 Liter of a 0.200 M acetic acid solution to m

Pie

<u>Answer:</u> The mass of sodium acetate that must be added is 30.23 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

$0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})$

$pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of acetic acid = 4.74

$[CH_3COONa]=?mol$

$[CH_3COOH]=0.200mol$

pH = 5.00

Putting values in above equation, we get:

$5=4.74+\log(\frac{[CH_3COONa]}{0.200})$

$[CH_3COONa]=0.364mol$

To calculate the mass of sodium acetate for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

$0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g$

Hence, the mass of sodium acetate that must be added is 30.23 grams

7 0

3 years ago

A student ran the following reaction in the laboratory at 311 K:CH4(g) + CCl4(g) 2CH2Cl2(g)When she introduced 4.10×10-2 moles o

statuscvo [17]

<u>Answer:</u> The value of $K_{eq}$ is 0.044

<u>Explanation:</u>

We are given:

Initial moles of methane = $4.10\times 10^{-2}mol=0.0410moles$

Initial moles of carbon tetrachloride = $6.51\times 10^{-2}mol=0.0651moles$

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of methane = $\frac{0.0410}{1.00}=0.0410M$

Concentration of carbon tetrachloride = $\frac{0.0651}{1.00}=0.0651M$

The given chemical equation follows:

$CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)$

<u>Initial:</u> 0.0410 0.0651

<u>At eqllm:</u> 0.0410-x 0.0651-x 2x

We are given:

Equilibrium concentration of carbon tetrachloride = $6.02\times 10^{-2}M=0.0602M$

Evaluating the value of 'x', we get:

$\Rightarrow (0.0651-x)=0.0602\\\\\Rightarrow x=0.0049M$

Now, equilibrium concentration of methane = $0.0410-x=[0.0410-0.0049]=0.0361M$

Equilibrium concentration of $CH_2Cl_2=2x=[2\times 0.0049]=0.0098M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[CH_2Cl_2]^2}{[CH_4]\times [CCl_4]}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.0098)^2}{0.0361\times 0.0603}\\\\K_{eq}=0.044$

Hence, the value of $K_{eq}$ is 0.044

8 0

3 years ago

How many mg of metallic ions are in 1.0 mL of 0.10 M LINO3

Anton [14]

Answer:

0.6941 mg

Explanation:

First we <u>calculate how many LiNO₃ moles there are</u>, using the <em>given concentration and volume</em>:

1.0 mL * 0.10 M = 0.10 mmol LiNO₃

As 1 mol of LiNO₃ contains 1 mol of Li,<em> in the problem solution there are 0.10 mmol of Li</em> (the only metallic ion present).

Now we<u> convert Li milimoles into miligrams</u>, using its <em>atomic mass</em>:

0.10 mmol Li * 6.941 mg/mmol = 0.6941 mg

8 0

3 years ago

Phosphorus trichloride, PCl₃ is a liquid at room temperature.

Simora [160]

Answer:

Phosphorus trichloride, PCl₃ undergoes change in bonding and molecular force of attraction, causing it to be liquid at room temperature.

Explanation:

Unlike other chlorides of Period 3 elements, Phosphorus trichloride, PCl₃ changes the structure of its molecular bonding from ionic to covalent bonds as it transitions to fluids (liquids or gases). The PCl₃ molecule also has the weak Van der Waals dispersion and dipole-dipole attraction, making it a fuming liquid at room temperature, with no electrical conductivity.

4 0

4 years ago

What is the ph of 0.450 m al(no3)3 [ka for al3+(aq) = 1.00x10-5]? express your answer to two decimal places?

White raven [17]

Answer : The pH of the solution is, 2.67

Explanation :

The equilibrium chemical reaction is:

$Al^{3+}+H_2O\rightarrow Al(OH)^{2+}+H^+$

Initial conc. 0.450 0 0

At eqm. (0.450-x) x x

As we are given:

$K_a=1.00\times 10^{-5}$

The expression for equilibrium constant is:

$K_a=\frac{(x)\times (x)}{(0.450-x)}$

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

$1.00\times 10^{-5}=\frac{(x)\times (x)}{(0.450-x)}$

$x=0.00212M$

The concentration of $H^+$ = x = 0.00212 M

Now we have to calculate the pH of solution.

$pH=-\log [H^+]$

$pH=-\log (0.00212)$

$pH=2.67$

Therefore, the pH of the solution is, 2.67

8 0

4 years ago