Question

1.00 mL of 12.0 M HCl is added to 1.00 L of a buffer that is 0.120 M HNO2 and 0.180 M NaNO2. How many moles of HNO2 and NaNO2 remain in solution after addition of the HCl?

Answer 1

Number of moles of NaNO2 = Volume of solution(in L) * Molarity of NaNO2 = 1L * 0.180 = 0.180 moles

Number of moles of HNO2 = Volume of solution(in L) * Molarity of HNO2 = 1L * 0.120 = 0.120 moles

Number of moles of HCl added = Volume of HCl * Molarity of HCl = 1/1000 * 12M = 0.012 moles

Added HCl will react with NaNO2 to form

HCl + NaNO2 ------ NaCl + HNO2

1 mole of HCl will react with 1 mole of NaNO2, so 0.012 moles of HCl will react with 0.012 moles of NaNO2

Moles of NaNO2 remained in solution = initial moles - number of moles of HCl = 0.180 - 0.012 = 0.168 moles

Moles of HNO2 remained in solution = initial moles + number of moles of HCl = 0.120 + 0.012 = 0.132 moles