Answer to Question #92725 in Chemistry for Tj

Question

A 25.00 mL sample of 0.379 M nitric acid (HNO3(aq)) is titrated with 12.00 mL of 0.573 M lithium hydroxide (LiOH(aq)). Determine the amount of [H+(aq)] remaining in the flask and pH of the solution after reaction.

Answer 1

Solution.

$LiOH + HNO3 = LiNO3 + H2O$

$OH^{-} + H^{+} = H2O$

$C = \frac{n}{V}$

$n(LiOH) = C(LiOH) \times V(LiOH)$

$n(HNO3) = C(HNO3) \times V(HNO3)$

n(LiOH) = 0.006876 mole

n(HNO3) = 0.009475 mole

n(LiOH)<n(HNO3), therefore lithium hydroxide will be limiting reagent.

Unconsumed nitric acid remained in the solution, so pH will be less than 7.

$n(HNO3 (left)) = 0.009475-0.006876 = 0.002599 mole$

$[H^{+}(aq)] = n(HNO3 (left)) = 0.002599 mole$

$pH = -lg([n(H^{+})])$

pH = 2.59

Answer:

$[H^{+} (aq)] = 0.002599 mole$

$pH = 2.59$