Q). What are the pH and H+ ion concentrations of a solution obtained by adding 0.8g of sodium hydroxide (Mr= 40) and 2.7g acetic acid (Mr=60) to 500mL water?

A). NaOH moles= 0.8/40=20

Acetic acid moles= 2.7/60=45

But then in the solution it says that NaOH conc in 500mL is 40 and acetic acid conc in 500mL is 90? How is this the case?

A). NaOH moles= 0.8/40=20

Acetic acid moles= 2.7/60=45

But then in the solution it says that NaOH conc in 500mL is 40 and acetic acid conc in 500mL is 90? How is this the case?

Original post by pigeonwarrior

Q). What are the pH and H+ ion concentrations of a solution obtained by adding 0.8g of sodium hydroxide (Mr= 40) and 2.7g acetic acid (Mr=60) to 500mL water?

A). NaOH moles= 0.8/40=20

Acetic acid moles= 2.7/60=45

But then in the solution it says that NaOH conc in 500mL is 40 and acetic acid conc in 500mL is 90? How is this the case?

A). NaOH moles= 0.8/40=20

Acetic acid moles= 2.7/60=45

But then in the solution it says that NaOH conc in 500mL is 40 and acetic acid conc in 500mL is 90? How is this the case?

There are definitely typos there - 0.8 g of NaOH is 0.02 mol and 2.7 g of CH3COOH is 0.045 mol. Correct digits, wrong position of the decimal place.

500 ml is 0.5 dm^3. Remember the formula for calculating a concentration using moles and volume?

Original post by TypicalNerd

There are definitely typos there - 0.8 g of NaOH is 0.02 mol and 2.7 g of CH3COOH is 0.045 mol. Correct digits, wrong position of the decimal place.

500 ml is 0.5 dm^3. Remember the formula for calculating a concentration using moles and volume?

500 ml is 0.5 dm^3. Remember the formula for calculating a concentration using moles and volume?

Thank you! 😀I understand that now. Here in the same question, I am stuck on the highlighted part. How come the total acetate produced was 40? And there was 50 remaining?

(edited 1 month ago)

Original post by pigeonwarrior

Thank you! 😀I understand that now. Here in the same question, I am stuck on the highlighted part. How come the total acetate produced was 40? And there was 50 remaining?

I hate the way the model solution is given - you can’t treat concentrations like amounts in most cases and so it’s better to work with moles and calculate concentrations (edited yet again: not moles - freudian slip lol) in the very last step.

Recall we said there were 0.02 mol (or 20 mmol) of NaOH and 0.045 mol (or 45 mmol) of CH3COOH? Let us write out the equation:

CH3COOH + NaOH —> CH3COONa + H2O

NaOH and CH3COOH react in a 1:1 molar ratio, meaning that all 20 mmol of NaOH will be used up, so 20 mmol of CH3COOH will be consumed as well, leaving just 25 mmol of the original 45 mmol. Since there are 25 mmol of acid left. This means the concentration is 50 mM since this amount is all dissolved in 0.5 dm^3 of water.

We can also see from the equation that for every mole of NaOH used up, a mole of CH3COONa is made, so 20 mmol of it is formed and as this is dissolved in that same 0.5 dm^3 of water, the concentration is (20/0.5) = 40 mM.

(edited 1 month ago)

Original post by TypicalNerd

I hate the way the model solution is given - you can’t treat concentrations like amounts in most cases and so it’s better to work with moles and calculate moles in the very last step.

Recall we said there were 0.02 mol (or 20 mmol) of NaOH and 0.045 mol (or 45 mmol) of CH3COOH? Let us write out the equation:

CH3COOH + NaOH —> CH3COONa + H2O

NaOH and CH3COOH react in a 1:1 molar ratio, meaning that all 20 mmol of NaOH will be used up, so 20 mmol of CH3COOH will be consumed as well, leaving just 25 mmol of the original 45 mmol. Since there are 25 mmol of acid left. This means the concentration is 50 mM since this amount is all dissolved in 0.5 dm^3 of water.

We can also see from the equation that for every mole of NaOH used up, a mole of CH3COONa is made, so 20 mmol of it is formed and as this is dissolved in that same 0.5 dm^3 of water, the concentration is (20/0.5) = 40 mM.

Recall we said there were 0.02 mol (or 20 mmol) of NaOH and 0.045 mol (or 45 mmol) of CH3COOH? Let us write out the equation:

CH3COOH + NaOH —> CH3COONa + H2O

NaOH and CH3COOH react in a 1:1 molar ratio, meaning that all 20 mmol of NaOH will be used up, so 20 mmol of CH3COOH will be consumed as well, leaving just 25 mmol of the original 45 mmol. Since there are 25 mmol of acid left. This means the concentration is 50 mM since this amount is all dissolved in 0.5 dm^3 of water.

We can also see from the equation that for every mole of NaOH used up, a mole of CH3COONa is made, so 20 mmol of it is formed and as this is dissolved in that same 0.5 dm^3 of water, the concentration is (20/0.5) = 40 mM.

thank you so so much for your help! It is much clearer now 😁

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