1.00 mL of 12.0 M HCl is added to 1.00 L of a buffer that contains 0.110 M HNO2 and 0.170 M NaNO2. How many moles of HNO2 and NaNO2 remain in solution after addition of the HCl

Moles of NaNO2 = 0.158

Moles of HNO2 final = 0.098

Explanation:

Given

Moles of HCl = 12

Moles of HNO2 = 0.11

Moles of NaNO2 = 0.170

HCl +NaNO2 --> HNO2  + NaCl

1 mole of HCl react with one mole of NaNO2 to produce 1 mole of NaCl and 1 mole of HNO2

Moles of NaNO2 = 0.17 - 0.012 = 0.158

Moles of HNO2 final = 0.11 - 0.012 = 0.098