Answer:
1) The neutralization reaction will mot be completed.
2) pH = 0.6.
Explanation:
<em>1) Will the neutralization reaction be completed?</em>
For the neutralization reaction be completed; The no. of millimoles of the acid must be equal the no. of millimoles of the base.
The no. of millimoles of 125 mL of 2.0 M HCl = MV = (2.0 M)(125.0 mL) = 250.0 mmol.
The no. of millmoles of 175 mL of 1.0 M NaOH = MV = (1.0 M)(175.0 mL) = 175.0 mmol.
∴ HCl will be in excess.
∴ The neutralization reaction will mot be completed.
<em>2) If not what is the pH of the final solution?</em>
[H⁺] =[ (MV)HCl - (MV)NaOH]/V total = (250.0 mmol - 175.0 mmol) / (300.0 mL) = 0.25 M.
∵ pH = - log[H⁺]
<em>∴ pH =</em> - log(0.25) =<em> 0.6.</em>
