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Chapter 1: Problem 25

The value of \(K_{a}\) for \(\mathrm{HSO}_{4}^{-}\) is \(1 \times 10^{-2} .\) What is the value of \(K_{b}\) for \(\mathrm{SO}_{4}^{2-} ?\) (A) \(K_{b}=1 \times 10^{-12}\) (B) \(K_{b}=1 \times 10^{-8}\) (C) \(K_{b}=1 \times 10^{-2}\) (D) \(K_{b}=1 \times 10^{2}\)

Short Answer

Expert verified
Therefore, the value of Kb for SO4^2- is 1x10^-12, which corresponds to option (A).

Step by step solution

01

Identifying Given Quantities

First, identify what the problem has provided: the Ka value for HSO4-, which is 1x10^-2.
02

Understanding the Relationship between Ka, Kb, and Kw

Next, understand the relationship between Ka, Kb, and Kw which is given by \(K_w = K_a \times K_b\). This is a crucial equation for tackling this problem. It means that the product of the Ka value of a specific acid and the Kb value of its conjugate base equals Kw.
03

Substituting the Values into the Equation

Now, substitute Ka into the aforementioned expression. We know that \(K_w = 1 x 10^{-14}\) at room temperature. Therefore, \(K_b = \frac{K_w}{K_a} = \frac{1 x 10^{-14}}{1 x 10^{-2}}\).
04

Simplifying the Expression

Finally, perform the division to find the value of Kb. Simplifying the expression will give you \(K_{b}=1 \times 10^{-12}\).

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Most popular questions from this chapter

When calcium chloride \(\left(\mathrm{CaCl}_{2}\right)\) dissolves in water, the temperature of the water increases dramatically. Compared to \(\mathrm{CaCl}_{2},\) what must be true regarding the hydration energy of \(\mathrm{CaF}_{2} ?\) (A) It would be greater because fluoride is smaller than chloride. (B) It would be the same because the charges of fluoride and chloride are identical. (C) It would be the same because hydration energy is only dependent on the IMFs present in water. (D) It would be smaller because the molar mass of \(\mathrm{CaF}_{2}\) is smaller than that of \(\mathrm{CaCl}_{2}\) .

\(\mathrm{CaCO}_{3}(s)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) 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

The following mechanism is proposed for a reaction: \(\begin{array}{ll}{2 \mathrm{A} \rightarrow \mathrm{B}} & {\text { (fast equilibrium) }} \\ {\mathrm{C}+\mathrm{B} \rightarrow \mathrm{D}} & {\text { (slow) }} \\ {\mathrm{D}+\mathrm{A} \rightarrow \mathrm{E}} & {\text { (fast) }}\end{array}\) Which of the following is the correct rate law for compete reaction? (A) Rate \(=k[\mathrm{C}]^{2}[\mathrm{B}]\) (B) Rate \(=k\left[\mathrm{Cl}[\mathrm{A}]^{2}\right.\) (C) Rate \(=k[\mathrm{C}][\mathrm{A}]^{3}\) (D) Rate \(=k[\mathrm{D}][\mathrm{A}]\)

A sample of a hydrate of \(\mathrm{CuSO}_{4}\) with a mass of 250 grams was heated until all the water was removed. The sample was then weighed and found to have a mass of 160 grams. What is the formula for the hydrate? (A) \(\mathrm{CuSO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}\) (B) \(\mathrm{CuSO}_{4} \cdot \mathrm{TH}_{2} \mathrm{O}\) (C) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) (D) \(\mathrm{CuSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)

A 0.1-molar solution of which of the following acids will be the best conductor of electricity? (A) \(\mathrm{H}_{2} \mathrm{CO}_{3}\) (B) \(\mathrm{H}_{2} \mathrm{S}\) (C) \(\mathrm{HF}\) (D) \(\mathrm{HNO}_{3}\)
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