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Nutka1998 [239]

4 years ago

10

I need help asaaaaap

1 answer:

SVETLANKA909090 [29]4 years ago

8 0

Answer:

NOT APPLICABLE

Explanation:

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3 years ago

If you mix 50mL of 0.1 M TRIS acid with 60 mL of0.2 M<br> TRIS base, what will be the resulting pH?

Katyanochek1 [597]

<u>Answer:</u> The pH of resulting solution is 8.7

<u>Explanation:</u>

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

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

<u>For TRIS acid:</u>

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

$0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol$

<u>For TRIS base:</u>

Molarity of TRIS base solution = 0.2 M

Volume of solution = 60 mL

Putting values in above equation, we get:

$0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol$

Volume of solution = 50 + 60 = 110 mL = 0.11 L (Conversion factor: 1 L = 1000 mL)

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

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

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

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of TRIS acid = 8.3

$[\text{TRIS acid}]=\frac{0.005}{0.11}$

$[\text{TRIS base}]=\frac{0.012}{0.11}$

pH = ?

Putting values in above equation, we get:

$pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7$

Hence, the pH of resulting solution is 8.7

6 0

4 years ago