Answer:
The solution formed is acidic
pH = 0.87
Explanation:
The acid-base reaction of NaOH with HCl is:
NaOH + HCl → NaCl + H₂O
Where 1 mole of NaOH reacts with 1 mole of HCl
In the problem, moles of NaOH and HCl are:
NaOH: 0.0250L × (0.525mol / L) = 0.013125 moles NaOH
HCl: 0.0750 × (0.355mol / L) = 0.026625 moles HCl
As moles of HCl > moles NaOH, HCl is in excess and the solution formed is acidic
Moles in excess of HCl are:
0.026625 moles - 0.013125 moles = 0.0135 moles HCl
As the volume of the solutions is 25.0mL + 75.0mL = 100.0mL = 0.100L, molarity of HCl after reaction is:
0.0135 moles HCl / 0.100L = 0.135M HCl = 0.135M H⁺
As pH is defined as - log [H⁺], pH of the solution is:
pH = -log 0.135M H⁺ = 0.87
pH = 0.87
The pH of the resulting solution of the acid and base mixture is; Acidic with a PH = 0.87
We are given;
Volume of NaOH = 25 mL = 0.025 L
Volume of HCl = 75 mL = 0.075 L
Concentration of NaOH = 0.525 M
Concentration of HCl = 0.355 M
Total volume of solution = 0.025 + 0.075 = 0.1 L
Now, the the equation of the reaction is;
HCl + NaOH = NaCl + H2O
We can see that 1 mole of HCl reacts with 1 mole of NaOH.
Since; number of moles = Volume × concentration
Number of moles of HCl = 0.075 × 0.355
Number of moles of HCl = 0.026625 moles
Similarly;
Number of moles of NaOH = 0.025 × 0.525
Number of moles of NaOH = 0.013125 moles
We can see that the number of moles of HCl is greater than that of NaOH and as such there is excess HCl acid in the mix.
Thus,
Excess moles of HCl acid = 0.026625 - 0.013125
Excess moles of HCl acid = 0.0135 moles
Concentration of this excess HCl is gotten by the formula;
M = Excess moles/total volume
M = 0.0135/0.1
M = 0.135 M
PH of this excess concentration is;
PH = -log [H+]
PH = -log [0.135]
PH = 0.87
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