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Chapter 1: Problem 2

A student titrates 20.0 \(\mathrm{mL}\) of 1.0 \(M \mathrm{NaOH}\) with 2.0 \(\mathrm{M}\), \(\mathrm{HCO}_{2} \mathrm{H}\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right) .\) Formic acid is a monoprotic acid. At the equivalence point, is the solution acidic, basic, or neutral? Why? (A) Acidic; the strong acid dissociates more than the weak base (B) Basic; the only ion present at equilibrium is the conjugate base (C) Basic; the higher concentration of the base is the determining factor (D) Neutral; equal moles of both acid and base are present

Short Answer

Expert verified
(B) Basic; the only ion present at equilibrium is the conjugate base

Step by step solution

01

Understanding the given

In the given problem, a base \(1.0 M \mathrm{NaOH}\) is being titrated with an acid \(\mathrm{HCO}_{2} \mathrm{H}\) which has a given \(K_{a}\) value (indicating it is a weak acid). The base and acid have specified volumes (20.0 mL and we'll find out for acid).
02

Identifying the equivalence point

At the equivalence point, the number of moles of the base is equal to the number of moles of the acid.
03

Calculating the moles of base

Calculate the quantity in moles of NaOH using the formula n = M×V, where n is the number of moles, M is molarity (concentration) and V is the volume. Here, M = 1.0 and the volume is 20 mL (we should convert this to liters = 0.020 L). We get: n = 1.0 × 0.020 = 0.020 moles.
04

Determining the nature of the solution

In this case, as the acid gets completely titrated by the base, it will donate protons to form carbonic acid (H2CO3) leaving the conjugate base \(HC0_3 ^{-}\) in the solution. Because the acid is weak and doesn't completely ionize and the base is strong and ionizes completely (forming OH- ions), the solution will have excess OH- ions making it basic.

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Most popular questions from this chapter

A bottle of water is left outside early in the morning. The bottle warms gradually over the course of the day. What will happen to the pH of the water as the bottle warms? (A) Nothing; pure water always has a pH of 7.00. (B) Nothing; the volume would have to change in order for any ion concentration to change. (C) It will increase because the concentration of \(\left[\mathrm{H}^{+}\right]\) is increasing. (D) It will decrease because the auto-ionization of water is an endothermic process.

\(2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{AlCl}_{3}(s)\) The reaction above is not thermodynamically favored under standard conditions, but it becomes thermodynamically favored as the temperature decreases toward absolute zero. Which of the following is true at standard conditions? (A) \(\Delta S\) and \(\Delta H\) are both negative. (B) \(\Delta S\) and \(\Delta H\) are both positive. (C) \(\Delta S\) is negative, and \(\Delta H\) is positive. (D) \(\Delta S\) is positive, and \(\Delta H\) is negative.

After 44 minutes, a sample of \(_{19}^{44} \mathrm{K}\) is found to have decayed to 25 percent of the original amount present. What is the half-life of \(_{19}^{44} \mathrm{K}?\) (A) 11 minutes (B) 22 minutes (C) 44 minutes (D) 66 minutes

In an experiment 2 moles of \(\mathrm{H}_{2}(g)\) and 1 mole of \(\mathrm{O}_{2}(g)\) were completely reacted, according to the following equation in a sealed container of constant volume and temperature: $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)$$ If the initial pressure in the container before the reaction is denoted as \(P_{i}\) which of the following expressions gives the final pressure, assuming ideal gas behavior? (A) \(P_{i}\) (B) 2\(P_{i}\) (C) \((3 / 2) P_{i}\) (D) \((2 / 3) P_{i}\)

\(\mathrm{CaCO}_{3}(s)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) If the reaction above took place at standard temperature and pressure and 150 grams of \(\mathrm{CaCO}_{3}(\mathrm{s})\) were consumed, what was the volume of \(\mathrm{CO}_{2}(g)\) produced at STP? (A) 11 L (B) 22 L (C) 34 L (D) 45 L
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