Question

asked May 4, 2020 179k views

1 Answer

Frank Wong · Answer 1 · 2020-05-08T11:16:19+0000

Answer: The new pH of resulting solution is 6.03

Step-by-step explanation:

We are adding hydrochloric acid to the solution, so it will react with salt (sodium hydrogen carbonate) only.

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

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

Molarity of hydrochloric acid = 0.15 M

Volume of solution = 3 mL

Putting values in above equation, we get:

$0.15M=\frac{\text{Moles of hydrochloric acid}* 1000}{3mL}\\\\\text{Moles of hydrochloric acid}=0.00045mol$

Molarity of sodium hydrogen carbonate = 0.025 M

Volume of solution = 150 mL

Putting values in above equation, we get:

$0.025M=\frac{\text{Moles of sodium hydrogen carbonate}* 1000}{150mL}\\\\\text{Moles of sodium hydrogen carbonate}=0.00375mol$

Molarity of carbonic acid = 0.045 M

Volume of solution = 150 mL

Putting values in above equation, we get:

$0.045M=\frac{\text{Moles of carbonic acid}* 1000}{150mL}\\\\\text{Moles of carbonic acid}=0.00675mol$

The chemical reaction for sodium hydrogen carbonate and hydrochloric acid follows the equation:

$NaHCO_3+HCl\rightarrow NaCl+H_2CO_3$

Initial: 0.00375 0.00045 0.00675

Final: 0.00330 - 0.00720

Volume of solution = 150 + 3 = 153 mL = 0.153 L (Conversion factor: 1 L = 1000 mL)

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

$pH=pK_a+\log(([salt])/([acid]))$

$pH=pK_a+\log(([NaHCO_3])/([H_2CO_3]))$

We are given:

pK_a = negative logarithm of acid dissociation constant of carbonic acid = 6.37

[NaHCO_3]=(0.0033)/(0.153)

[H_2CO_3]=(0.0072)/(0.153)

pH = ?

Putting values in above equation, we get:

$pH=6.37+\log((0.0033/0.153)/(0.0072/0.153))\\\\pH=6.03$

Hence, the new pH of the solution is 6.03

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