Question

Calculate the pH of a solution formed by mixing 20.00 mL of 0.30 M HCl with 30.00 mL of 0.15 M HNO3??

Answer 1

0.68

Step-by-step explanation:

Both
`HCl` and
`HNO_3` are strong acids which ionize completely in water. According to stoichiometry, 1 mole of each will produce 1 mole of hydronium cations. Therefore, let's calculate the total number of moles of hydronium and find the final concentration of it keeping in mind the dilution:

`n_(H_3O^+) = n_(HCl) + n_(HNO_3) = c_(HCl)V_(HCl) + c_(HNO_3)V_(HNO_3)`

Total molarity:

`[H_3O^+] = (c_(HCl)V_(HCl) + c_(HNO_3)V_(HNO_3))/(V_(HCl) + V_(HNO_3))`

Find pH:

`pH = -\log[H_3O^+] = -\log((c_(HCl)V_(HCl) + c_(HNO_3)V_(HNO_3))/(V_(HCl) + V_(HNO_3))) = -\log((0.30~M\cdot 20.00~mL + 0.15~M\cdot 30.00~mL)/(20.00~mL + 30.00~mL)) = 0.68`