What is the normality of 10 ml of an acid that took 25 ml of 0.25 N NaOH to reach pH 7?
It depends upon which acid is being titrated with NaOH because the equivalence point is NOT always at 7.0. However, again, I assume this is a simple case. Use mL(acid) x N(acid) = mL(NaOH) x N(NaOH). The only unknown is N(acid).
To determine the normality (N) of the acid, we need to use the neutralization reaction between the acid and NaOH. The balanced equation for the neutralization reaction is:
Acid + NaOH → Salt + Water
From the given information, we know that 25 ml of 0.25 N NaOH (sodium hydroxide) was required to reach pH 7. This means that the acid and NaOH reacted in a 1:1 molar ratio.
Now, let's calculate the amount of NaOH (in moles) used in this reaction:
Amount of NaOH (in moles) = Concentration of NaOH (N) × Volume of NaOH (in liters)
= 0.25 N × 0.025 L
= 0.00625 moles
Since the acid and NaOH react in a 1:1 ratio, the amount of acid used is also 0.00625 moles.
Now, let's find the normality (N) of the acid:
Normality (N) = Amount of acid (in moles) / Volume of acid (in liters)
= 0.00625 moles / 0.010 L
= 0.625 N
Therefore, the normality of the acid in this reaction is 0.625 N.