40,0 mL of 0,125 M Mg(OH)2 and 150,0 mL of 0,125 M HCl is mixed together. What is the pH of this solution?
we have:
Molarity of H+ = 0.125 M
Volume of H+ = 150 mL
Molarity of OH- = 2*[Mg(OH)2] = 2*0.125 M = 0.250 M
Volume of OH- = 40 mL
mol of H+ = Molarity of H+ * Volume of H+
mol of H+ = 0.125 M * 150 mL = 18.75 mmol
mol of OH- = Molarity of OH- * Volume of OH-
mol of OH- = 0.25 M * 40 mL = 10 mmol
We have:
mol(H+) = 18.75 mmol
mol(OH-) = 10 mmol
10 mmol of both will react
remaining mol of H+ = 8.75 mmol
Total volume = 190.0 mL
[H+]= mol of acid remaining / volume
[H+] = 8.75 mmol/190.0 mL
= 0.0461 M
we have below equation to be used:
pH = -log [H+]
= -log (4.605*10^-2)
= 1.34
Answer: 1.34
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