Which of the following solutions should be used when titrating a 25.00mL sample of CH3COOH that is approximately 0.1 M?
a.) 0.150 M NaOH
b.) 0.001 M NaOH
c.) 3.00 M NaOH
d.) 6.00 M NaOH
Well, if you want to have some fun with your titration, I would suggest using option b.) 0.001 M NaOH. It's always exciting to see if you can spot that subtle color change when adding such a tiny amount of titrant. Just make sure you have good eyesight or a magnifying glass ready!
To determine the most appropriate solution for titrating the CH3COOH sample, we need to consider the reaction that occurs during titration.
CH3COOH (acetic acid) reacts with NaOH (sodium hydroxide) in a 1:1 stoichiometric ratio:
CH3COOH + NaOH -> CH3COONa + H2O
During titration, the goal is to neutralize the acetic acid by adding a base (NaOH) until the equivalence point is reached.
The equivalence point is achieved when all of the acetic acid has reacted with the NaOH, resulting in a stoichiometrically equivalent amount of NaOH and CH3COOH.
To effectively reach the equivalence point, it is important to choose a solution of NaOH with a concentration close to the molarity of CH3COOH in the sample.
In this case, the sample of CH3COOH is approximately 0.1 M.
Among the given options, the best choice would be a solution with a concentration closest to 0.1 M:
a) 0.150 M NaOH
This option is the closest in concentration to the 0.1 M acetic acid solution and would be the most suitable for titrating the CH3COOH sample.
To determine the appropriate solution for titrating a sample of CH3COOH, you need to consider the stoichiometry of the reaction between CH3COOH and NaOH. The balanced equation for the reaction is:
CH3COOH + NaOH -> CH3COONa + H2O
From the equation, we can determine that the molar ratio between CH3COOH and NaOH is 1:1. This means that for every mole of CH3COOH, we need one mole of NaOH to reach the equivalence point.
First, let's calculate the number of moles of CH3COOH in the 25.00 mL sample. Given that the concentration is approximately 0.1 M, we can use the formula:
moles = concentration * volume
moles of CH3COOH = 0.1 M * 0.025 L
moles of CH3COOH = 0.0025 moles
Now that we know the number of moles of CH3COOH, we need to find a solution of NaOH with the same number of moles or higher to ensure complete reaction. Let's analyze each option:
a.) 0.150 M NaOH: To calculate the number of moles of NaOH in 25.00 mL, we use the formula:
moles = concentration * volume
moles of NaOH = 0.150 M * 0.025 L
moles of NaOH = 0.00375 moles
This option provides a higher number of moles of NaOH than CH3COOH, which would ensure complete reaction. So, option a) can be used.
b.) 0.001 M NaOH: Using the same calculation, we find that this option provides a significantly lower number of moles of NaOH (0.000025 moles) than the moles of CH3COOH (0.0025 moles). Therefore, option b) is not suitable.
c.) 3.00 M NaOH: Calculating the number of moles of NaOH:
moles of NaOH = 3.00 M * 0.025 L
moles of NaOH = 0.075 moles
This option provides a much higher number of moles of NaOH than CH3COOH, which would result in an excessive amount of NaOH. Therefore, option c) is not suitable.
d.) 6.00 M NaOH: Since the concentration of NaOH in this option is even higher than option c), it would provide an even greater excess of NaOH. Hence, option d) is not suitable.
In conclusion, the appropriate solution to use for titrating the 25.00 mL sample of CH3COOH is option a) 0.150 M NaOH.