Answer to Question #244953 in General Chemistry for AsH

Question

1.00 mL of 12.0 M HCl is added to 1.00 L of a buffer that contains  0.110 M HNO₂ and 0.150 M NaNO₂. How many moles of HNO₂ and NaNO₂remain in solution after addition of the HCl?

Answer 1

Number of moles of NaNO2 = $Volume \;of \;solution(in \;L) \times \;Molarity \;of\; NaNO2 = 1L \times 0.150 = 0.150 \;moles$

Number of moles of HNO2 = $Volume\; of\; solution(in L) \times \;Molarity\; of \;HNO2 = 1L \times 0.110 = 0.110\; moles$

Number of moles of HCl added = $Volume \;of \;HCl \times Molarity\; of \;HCl = \frac{1}{1000} \times12M = 0.012 \;moles$

Added HCl will react with NaNO2 to form

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

Moles of NaNO2 remained in solution = initial moles - number of moles of HCl = 0.150 - 0.012 = 0.138 moles

Moles of HNO2 remained in solution = initial moles + number of moles of HCl = 0.110 + 0.012 = 0.122 moles