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 Na NO2 remain in solution after addition of the HCl
1 answer:
Answer:
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
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