A solution is made by mixing \(1.5 \mathrm{~g}\) of \(\mathrm{LiOH}\) and $23.5 \mathrm{~mL}\( of \)1.000 \mathrm{M}$ HNO3. (a) Write a balanced equation for the reaction that occurs between the solutes. (b) Calculate the concentration of each ion remaining in solution. (c) Is the resulting solution acidic or basic?

Short Answer

Expert verified
The balanced chemical equation for the reaction between LiOH and HNO3 is: \( LiOH + HNO_3 \rightarrow LiNO_3 + H_2O \). After the reaction, there are 1.668 M OH- ions, and 1.000 M Li+ and NO3- ions remaining in the solution. As there are no H+ ions remaining and the presence of OH- ions, the solution is basic.

Step by step solution

01

Write the balanced chemical equation

The reaction between LiOH (a base) and HNO3 (an acid) is an acid-base neutralization reaction, which produces water and a salt. The balanced chemical equation for the reaction is: \( LiOH + HNO_3 \rightarrow LiNO_3 + H_2O \)
02

Convert given amount of reactants into moles

As we are given mass of LiOH and the volume and concentration of HNO3, we have to convert these amounts into moles. Given, mass of LiOH = 1.5 g Molar mass of LiOH = 6.94 (Li) + 15.999 (O) + 1.008 (H) = 23.942 g/mol Moles of LiOH = mass/molar mass Moles of LiOH = 1.5 g / 23.942 g/mol = 0.0627 mol Given, volume of HNO3 = 23.5 mL = 0.0235 L Concentration of HNO3 = 1.000 M Moles of HNO3 = concentration × volume Moles of HNO3 = 1.000 M × 0.0235 L = 0.0235 mol
03

Determine the limiting reactant

Let's compare the mole ratio of reactants based on the balanced equation and the mole ratio from the given amounts of reactants. From the balanced equation: 1 mol LiOH : 1 mol HNO3 Given amounts: 0.0627 mol LiOH : 0.0235 mol HNO3 Since the given amounts of LiOH : HNO3 are greater than the mole ratio in the balanced equation, HNO3 is the limiting reactant. The reaction will consume all of the HNO3 and only a portion of the LiOH.
04

Calculate the remaining amounts of reactants and products

As HNO3 is the limiting reactant, it will be completely consumed. This means that in the resulting solution, there will be no HNO3 left and no H+ ions (from the consumed HNO3). Moles of LiOH remaining = Initial moles of LiOH - moles of consumed LiOH Moles of LiOH remaining = 0.0627 mol - 0.0235 mol = 0.0392 mol Moles of LiNO3 produced = moles of HNO3 consumed = 0.0235 mol
05

Calculate the concentration of each remaining ion

The total volume of the solution (assuming it does not change significantly due to the reaction) = 23.5 mL = 0.0235 L Concentration of remaining OH- ions (from LiOH) = moles of LiOH remaining / total volume Concentration of OH- ions = 0.0392 mol / 0.0235 L = 1.668 M Concentration of remaining Li+ and NO3- ions (from LiNO3) = moles of LiNO3 produced / total volume Concentration of Li+ and NO3- ions = 0.0235 mol / 0.0235 L = 1.000 M
06

Determine whether the solution is acidic or basic

Since there are no H+ ions remaining in the solution but there are OH- ions present, the solution will be basic. This is because a higher concentration of OH- ions compared to H+ ions will result in a pH greater than 7.

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