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3 years ago

Salts as neutral, acidic, or basic.? identify the following salts as neutral, acidic, or basic. 2srbr2 4lif

1 answer:

8 0

Acidic salts are those that are products of the reaction between a strong acid and a weak base, while basic salts are produced from reactions between weak acids and strong bases. In neutral salts, the strength of the reacting acid and base are similar.
SrBr₂ is a neutral salt
LiF is a basic salt

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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]

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

<u>Explanation:</u>

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

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