Chemistry, 14.09.2019 08:30 espinosajoselyn

Question 8: a 150 ml of buffer contains 0.045 m carbonic acid (h2co3) and 0.025 m sodium hydrogen carbonate (hco3). if 3.0 ml of 0.15m hcl is added to the buffer, what is the new ph? pka = 6.37

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Answer from: borsha255

The new pH of resulting solution is 6.03

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:

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

For hydrochloric acid:

Molarity of hydrochloric acid = 0.15 M

Volume of solution = 3 mL

Putting values in above equation, we get:

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

For sodium hydrogen carbonate:

Molarity of sodium hydrogen carbonate = 0.025 M

Volume of solution = 150 mL

Putting values in above equation, we get:

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

For carbonic acid:

Molarity of carbonic acid = 0.045 M

Volume of solution = 150 mL

Putting values in above equation, we get:

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

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:

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

$pH=pK_a+\log(\frac{[NaHCO_3]}{[H_2CO_3]})$

We are given:

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

$[NaHCO_3]=\frac{0.0033}{0.153}$

$[H_2CO_3]=\frac{0.0072}{0.153}$

pH = ?

Putting values in above equation, we get:

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

Hence, the new pH of the solution is 6.03

Answer from: Quest

The correct answer is "atomic weight".

Answer from: Quest

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o

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Question 8: a 150 ml of buffer contains 0.045 m carbonic acid (h2co3) and 0.025 m sodium hydrogen c...

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