The number of electrons in the valence shell determine what types of bonds an atom can form.

Suppose that you have 115 mL of a buffer that is 0.460 M in both benzoic acid ( C 6 H 5 COOH ) and its conjugate base ( C 6 H 5

77julia77 [94]

Answer: The maximum volume of HCl that can be added before its buffering capacity is lost is 111.3 mL

Explanation:

We are given:

Concentration of buffer = 0.460 M

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

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

$pH=pK_a+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})$ .....(1)

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of benzoic acid = 4.2

$[C_6H_5COOH]=0.460M$

$[C_6H_5COO^-]=0.460M$

pH = ?

Putting values in equation 1, we get:

$pH=4.2+\log(\frac{0.460}{0.460})\\\\pH=4.2$

When the pH of the buffer changes by 1 unit, the buffering capacity is said to be lost.

pH change for loosing buffer capacity = [4.2 - 1] = 3.2

Calculating the ratio of conjugate base and its acid by using equation 1:

$pK_a$ = 4.2

pH = 3.2

Putting values in equation 1, we get:

$3.2=4.2+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})\\\\\frac{[C_6H_5COO^-]}{[C_6H_5COOH]}=0.1$

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

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ ......(2)

For benzoic acid and its conjugate base:

Molarity of benzoic acid and its conjugate base = 0.460 M

Volume of solution = 115 mL

Putting values in equation 2, we get:

$0.460=\frac{\text{Moles of benzoic acid and its conjugate base}\times 1000}{115mL}\\\\\text{Moles of benzoic acid and its conjugate base}=\frac{0.460\times 115}{1000}=0.053mol$

The chemical reaction for aniline and HCl follows the equation:

$C_6H_5COO^-+HCl\rightarrow C_6H_5COOH+Cl^-$

Let the moles of acid added to carry out the change is 'x' moles

Calculating the moles of acid added:

$\frac{[C_6H_5COO^-]-x}{[C_6H_5COOH]+x}=0.1\\\\\frac{0.053-x}{0.053+x}=0.1\\\\x=0.0434$

Calculating the volume of acid added by using equation 2, we get:

Moles of acid added = 0.0434 moles

Molarity of solution = 0.390 M

Putting values in equation 2, we get:

$0.390M=\frac{0.0434\times 1000}{\text{Volume of acid}}\\\\\text{Volume of acid}=\frac{0.0434\times 1000}{0.390}=111.3mL$

Hence, the maximum volume of HCl that can be added before its buffering capacity is lost is 111.3 mL

7 0

3 years ago

The country like Nepal with diversity must have competition and cooperation justify

Len [333]

Help????? You need to write as if you are Dmitri Mendeleev and you have just made your proposed version of the periodic table. You need to write to the Royal Society of Chemistry in London explaining:
• Who you are
• What your periodic table is like, the groups, elements and features
• Why you think your periodic table is correct
• How you have built on the work of others or why you think their work is not
correct

3 0

3 years ago

Who ever answer this is genius ????????? let's see?

Flura [38]

Can you do plsssssssss

help me plssssssssssss meee

3 0

3 years ago

According to the vsepr model the arrangement of electron pairs around nh3 and ch4 is

bazaltina [42]

The arrangement of electron pairs around CH4 and NH3, According to the VSEPR model is the same, because in each case there are the same number of electron pairs around the central atom. So the NH3 and CH4 arrangement of electron pairs is the same because in each case there are the same number of electron pairs around the central atom.

7 0

3 years ago

A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o

julsineya [31]

Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

$Number \, of \, moles, n = \frac{Mass}{Molar \ mass} = \frac{5.84}{58.44} = 0.09993 \ moles \ of \ NaCl$

0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

3 0

3 years ago