A 25 mL sample of wine is found to have a concentration of acetic acid of 0.23 M. If this was titrated with a 0.16 M NaOH solution how many mL's of NaOH would be required?
How many moles of acetic acid do you have. M x L = ??
How many moles NaOH will that require? It must be the same because the reaction between acetic acid and NaOH is 1:1.
Then moles NaOH = M x L; you have moles NaOH and M NaOH, solve for L.
For 1:1 reactions, the shorter mL x M = mL x M may be used.
To find out how many milliliters (mL) of NaOH would be required to titrate the wine sample, we need to use the concept of stoichiometry. This involves using the balanced chemical equation and the molar ratios between the substances involved in the reaction.
Given:
Volume of wine sample (acetic acid solution) = 25 mL
Concentration of acetic acid = 0.23 M
Concentration of NaOH solution = 0.16 M
The balanced chemical equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is:
CH3COOH + NaOH → CH3COONa + H2O
From the equation, we can see that the molar ratio between acetic acid and NaOH is 1:1. This means that for every 1 mole of acetic acid, we need 1 mole of NaOH to react completely.
To calculate the amount of acetic acid in moles, we use the formula:
Amount (in moles) = concentration × volume (in liters)
Converting the volume of the wine sample to liters:
Volume (in liters) = 25 mL / 1000 = 0.025 L
Calculating the amount of acetic acid in moles:
Amount of acetic acid (in moles) = 0.23 M × 0.025 L = 0.00575 moles
Since the molar ratio between acetic acid and NaOH is 1:1, we need an equal amount of NaOH to react completely. Therefore, the amount of NaOH required is also 0.00575 moles.
To find the volume of NaOH solution required, we rearrange the formula:
Volume (in liters) = amount (in moles) / concentration (in M)
Calculating the volume of NaOH solution required (in liters):
Volume (in liters) = 0.00575 moles / 0.16 M = 0.0359375 L
Finally, converting the volume to milliliters:
Volume (in mL) = 0.0359375 L × 1000 = 35.9375 mL
Therefore, approximately 35.94 mL of NaOH solution would be required to titrate the 25 mL sample of wine.