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

Answer:

0.68

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:

$[H_3O^+] = \frac{c_{HCl}V_{HCl} + c_{HNO_3}V_{HNO_3}}{V_{HCl} + V_{HNO_3}}$

Find pH:

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