physical science
it is found that 40 ml of a 0.5mol.dm^-3 sodium hydroxide solution is needed to neutralise 20mol of the vinegar. calculate the pH of the sodium hydroxide solution
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A weird question; i.e., all of that information about vinegar but the pH of the NaOH has nothing to do with it.
However, pOH = -log(OH^-) = -log(0.5M) = ?
Then pH + pOH = pKw = 14. You know pOH and pKw, solve for pH. I get about 13.7.
To calculate the pH of the sodium hydroxide solution, we need to understand the reaction that takes place between sodium hydroxide (NaOH) and vinegar (acetic acid, CH3COOH).
From the given information, we know that 40 ml (0.04 L) of a 0.5 mol.dm^-3 (0.5 mol/L) sodium hydroxide solution is required to neutralize 20 mol of vinegar.
The reaction between sodium hydroxide and acetic acid can be represented as follows:
CH3COOH + NaOH → CH3COONa + H2O
From the balanced equation, we can see that the ratio of acetic acid to sodium hydroxide is 1:1.
This means that 20 mol of acetic acid will react with 20 mol of sodium hydroxide to form 20 mol of sodium acetate (CH3COONa) and release 20 mol of water (H2O).
Now, let's calculate the number of moles of sodium hydroxide present in the given volume.
Number of moles of sodium hydroxide = concentration (mol.dm^-3) x volume (dm^3)
= 0.5 mol.dm^-3 x 0.04 dm^3
= 0.02 mol
Since the ratio of acetic acid to sodium hydroxide is 1:1, we can conclude that the number of moles of acetic acid present in the reaction is also 0.02 mol.
To find the pH of the sodium hydroxide solution, we need to convert the number of moles of sodium hydroxide into concentration. The equation relating the concentration (C) and number of moles (n) is:
C = n / V
where C is the concentration in mol.dm^-3, n is the number of moles, and V is the volume in dm^3.
In this case, the volume is given as 0.04 L, which is equivalent to 0.04 dm^3.
Concentration of sodium hydroxide = 0.02 mol / 0.04 dm^3
= 0.5 mol.dm^-3
Now, let's calculate the pOH (the negative logarithm of the hydroxide ion concentration) of the sodium hydroxide solution:
pOH = -log10(OH^- concentration)
Since sodium hydroxide is a strong base, it completely dissociates in water, so the concentration of OH^- ions is equal to the concentration of sodium hydroxide.
Using the concentration calculated previously, we find:
pOH = -log10(0.5)
= -(-0.3010) (using logarithm properties)
= 0.3010
Finally, we can calculate the pH (the negative logarithm of the hydrogen ion concentration) using the relationship:
pH + pOH = 14
pH + 0.3010 = 14
pH = 14 - 0.3010
= 13.699
Therefore, the pH of the sodium hydroxide solution is approximately 13.699.