In the acid-base reaction between Sr(OH)2 and HNO3, 0.75 L of .85 M Sr(NO3)2 is formed upon the reaction of 0.515 L of HNO3 solution and excess Sr(OH)2. What is the molarity of the initial HNO3 solution?
200 mL of 1.5 M HCl is mixed with 150 mL of 3.0 M NaOH.
HCl + NaOH --> NaCl + H2O
What is the molarity of the resulting salt solution?
We never went over this stuff in class so if you could explain your answer that would be great. Thank you!
200 mL of 1.5 M HCl is mixed with 150 mL of 3.0 M NaOH.
HCl + NaOH --> NaCl + H2O
What is the molarity of the resulting salt solution?
We never went over this stuff in class so if you could explain your answer that would be great. Thank you!
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Both of these problems are of the general form:
M2 = V1M1 / V2
Sr(OH)2 + 2 HNO3 → Sr(NO3)2 + 2 H2O
(0.75 L) x (0.85 M) x (2/1) / (0.515 L HNO3) = 2.48 M HNO3
The HCl is the limiting reactant, so supposing the volumes to be additive:
(200 mL) x (1.5 M HCl) x (1/1) / (200 mL + 150mL) = 0.857 M NaCl
M2 = V1M1 / V2
Sr(OH)2 + 2 HNO3 → Sr(NO3)2 + 2 H2O
(0.75 L) x (0.85 M) x (2/1) / (0.515 L HNO3) = 2.48 M HNO3
The HCl is the limiting reactant, so supposing the volumes to be additive:
(200 mL) x (1.5 M HCl) x (1/1) / (200 mL + 150mL) = 0.857 M NaCl