# chemistry question

25.00 cm3 of propanoic acid, with a concentration of 0.120 mol dm−3, was pipetted into a conical flask. This solution was titrated with sodium hydroxide of concentration 0.150 mol dm−3. CH3CH2COOH + NaOH o CH3CH2COONa + H2O

(iii) Calculate the minimum volume of sodium hydroxide required to react with all of the propanoic acid. (2)

ans=20cm^3

iv) Calculate the pH when 40 cm3

of sodium hydroxide (an excess) was added.

ans:-(Excess NaOH = 20.0 cm3 in total volume of 65

cm3 ) [OH− ] = (20.0 x 0.15)/65 = 0.046154 [H+ ] = 1 x 10^−14/0.46154 = 2.1667 x 10−13 pH = (−log 2.1667 x 10−13 ) = 12.6642/12.7

can anyone explain why we consider the unreacted naoh =20 cm3 in the calculation, wont the reacted 20cm3 affect the ph as well?
Original post by Prof Claire
In the given titration scenario, sodium hydroxide (NaOH) is added to propanoic acid (CH3CH2COOH). The key to understanding why we consider the unreacted NaOH in pH calculation lies in recognizing the excess NaOH after the reaction is complete.
Here's the breakdown:
(iii) *Minimum Volume of NaOH Required:*
- The balanced chemical equation shows a 1:1 ratio between propanoic acid (CH3CH2COOH) and sodium hydroxide (NaOH).
- Since the concentration of propanoic acid is given, you can use the formula $$\text{moles} = \text{concentration} \times \text{volume}$$ to find the moles of propanoic acid.
- Then, using the stoichiometry of the reaction, you find that the moles of NaOH required are the same.
- Finally, you use the concentration and moles of NaOH to find the minimum volume of NaOH needed.
(iv) *pH Calculation after Excess NaOH:*
- After adding 40 cm3 of NaOH (which is in excess), it means the reaction has gone to completion, and there is unreacted NaOH present.
- The remaining unreacted NaOH contributes to the overall pH of the solution.
- Excess NaOH means you have both the original NaOH and the NaOH that reacted with the propanoic acid. Both contribute to the hydroxide ion concentration [OH−].
- In the pH calculation, you consider the total volume of the solution, including the volume of the reacted NaOH (20 cm3) and the excess NaOH (20 cm3).
- The total concentration of OH− is calculated using the excess NaOH, and then the pH is determined using the formula $$\text{pH} = -\log[\text{H}^+]$$, where $$[\text{H}^+]$$ is derived from the $$1 \times 10^{-14} = [\text{H}^+][\text{OH}^-]$$ relation.
In summary, the unreacted NaOH is considered in the calculation because it contributes to the overall hydroxide ion concentration, which, in turn, affects the pH of the solution. Both the reacted and unreacted portions of NaOH play a role in determining the pH after the titration is complete.

This looks like an AI generated response...
Original post by hamza.s913
25.00 cm3 of propanoic acid, with a concentration of 0.120 mol dm−3, was pipetted into a conical flask. This solution was titrated with sodium hydroxide of concentration 0.150 mol dm−3. CH3CH2COOH + NaOH o CH3CH2COONa + H2O

(iii) Calculate the minimum volume of sodium hydroxide required to react with all of the propanoic acid. (2)

ans=20cm^3

iv) Calculate the pH when 40 cm3

of sodium hydroxide (an excess) was added.

ans:-(Excess NaOH = 20.0 cm3 in total volume of 65

cm3 ) [OH− ] = (20.0 x 0.15)/65 = 0.046154 [H+ ] = 1 x 10^−14/0.46154 = 2.1667 x 10−13 pH = (−log 2.1667 x 10−13 ) = 12.6642/12.7

can anyone explain why we consider the unreacted naoh =20 cm3 in the calculation, wont the reacted 20cm3 affect the ph as well?

As shown in the equation, the reaction between the NaOH and the propanoic acid produces CH3CH2COONa (a salt, CH3CH2COO- and Na+) and water you might know this as a neutralisation reaction. Clue's kind of in the name I guess.

A couple of chemguide pages on this (there are more pages on acids and bases available), may or may not be helpful:
https://www.chemguide.co.uk/basicorg/acidbase/acids.html
https://www.chemguide.co.uk/physical/acidbaseeqia/theories.html
Original post by bl0bf1sh
As shown in the equation, the reaction between the NaOH and the propanoic acid produces CH3CH2COONa (a salt, CH3CH2COO- and Na+) and water you might know this as a neutralisation reaction. Clue's kind of in the name I guess.

A couple of chemguide pages on this (there are more pages on acids and bases available), may or may not be helpful:
https://www.chemguide.co.uk/basicorg/acidbase/acids.html
https://www.chemguide.co.uk/physical/acidbaseeqia/theories.html

yes i do understand that but any idea why we dont use 40cm3 in our calculation and why we just use 20cm3 which is the unreacted amount
Original post by hamza.s913
yes i do understand that but any idea why we dont use 40cm3 in our calculation and why we just use 20cm3 which is the unreacted amount

40 cm3 is how much NaOH is added to the mixture. In part iii) you worked out that 20 cm3 of NaOH reacts with the propanoic acid, meaning that NaOH is 20 cm3 in excess.

Part iii) is just a calculation, the NaOH hasn't been added yet. The 40 cm3 of NaOH is added in part iv). 20 cm3 of this reacts with the propanoic acid, and the other 20 cm3 of it remains in the mixture unreacted.