Question

1.00 mL of 12.0 M HCl is added to 1.00 L of a buffer that is 0.100 M HNO2 and .190 M NaNO2. How many moles of HNO2 and NaNO2 remain in solution after of the HCl

Answer 1

Before adding HCl

Molarity of HNO2 = 0.1 M

Molarity of NaNO2 = 0.190 M

Volume of solution = 1.00 L

Moles of HNO2 = Molarity x volume = 0.1 x 1 L = 0.1 mol

Moles of NaNO2 = Molarity x volume = 0.190 x 1 L = 0.190 mol

After adding HCl

Volume of HCl = 1.00 mL = 1 x10^-3 L = 0.001 L

Moles of HCl = Molarity x volume = 12 x 1 x10^-3 = 0.012 mol HCl

NaNO₂ + HCl --------> HNO₂ + NaCl

Mole ratio of NaNO2 and HCl is in 1:1 mole ratio

Moles of NaNO2 reacted with 0.012 mol HCl =

= 0.0120 mol NaNO₂

Moles of NaNO₂ remained in the solution = 0.190 - 0.0120 mol = 0.178 mol NaNO₂

According to the reaction 0.012 moles of HNO₂ is newly formed

Total moles of HNO₂ in the solution = 0.1 mol + 0.012 mol = 0.112 mol