Question

1.00 mL of 12.0 M HCl is added to 1.00 L of a buffer that is 0.130 M HNO2 and 0.200 M NaNO2. How many moles of HNO2 and NaNO2 remain in solution after addition of the HCl?

moles HNO2 =

moles NaNO2 =

Answer 1

mol of HCl added = 12.0M *0.001 L = 0.012 mol

NaNO2 will react with H+ to form HNO2

Before Reaction:

mol of NaNO2 = 0.2 M *1.0 L

mol of NaNO2 = 0.2 mol

mol of HNO2 = 0.13 M *1.0 L

mol of HNO2 = 0.13 mol

after reaction,

mol of NaNO2 = mol present initially - mol added

mol of NaNO2 = (0.2 - 0.012) mol

mol of NaNO2 = 0.188 mol

mol of HNO2 = mol present initially + mol added

mol of HNO2 = (0.13 + 0.012) mol

mol of HNO2 = 0.142 mol

mol of HNO2 = 0.142 mol

mol of NaNO2 = 0.188 mol