Answer:
see explanation
Step-by-step explanation:
Given 0.100M NaOH + 50ml 0.100M HCl
If 50ml of 0.100M NaOH is added to 50ml 0.100M HCl => pH = 7
if less than 50ml of 0.100M NaOH => pH < 7
if more than 50ml of 0.100M NaOH => pH > 7.
Example:
50ml(0.100M NaOH) + 50ml(0.100MHCl) =>
0.500mole NaCl + 0.500mole H₂O
In this mix, both NaOH and HCl are converted to NaCl and H₂O and NaOH or HCl are no longer present giving a neutral solution. Since neither Na⁺ nor Cl⁻ undergo hydrolysis, the pH is dependent upon H₂O ⇄ H⁺ + OH⁻ and [H⁺] = [OH⁻] = 10⁻⁷M => pH = -log[H⁺] = -log(10⁻⁷) = -(-7) = 7
If less than 50ml of 0.100M NaOH is added, the mix will have an excess of HCl and the pH will be acidic; i.e., less than 7.
If more than 50ml of 0.100M NaOH is added, the mix will have an excess of NaOH and the pH will be alkaline (basic); i.e., greater than 7.