Chemistry, 06.10.2019 16:30 4presidents

Questions 3-6 refer to rhe solutions below:
(a)100ml of 1.0 m ch3cooh mixed with 100ml of 1.0 m ch3coona
(b)100ml of 1.0 m hbr mixed with 100ml of 1.0 m kbr
(c)100ml of 1.0 m hi mixed with 100ml of 1.0 m naoh
(d)100ml of 1.0 m nh4cl mixed with 100ml of 1.0 m nh3
(e)100ml of 1.0 m naoh mixed with 100ml of 1.0 m ch3cooh

3.) which solution will produce a buffer with a ph < 6.5
4.) which solution will produce a buffer with a ph > 7.5
5.) which solutiom will have a ph of about 7
6.) which solution will be the most acidic

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Answers

Answer from: sarahhN7534

Under room temperature where $\text{pK}_w = 14$ :

3.) (A), (B), and (E).

4.) (D).

5.) (B).

Explanation

What makes a buffer solution? For a solution to be a buffer, it needs to contain large amounts of a weak acid and its conjugate base ion. Alternatively, the solution may contain large amounts of a weak base and its conjugate acid ion.

Not every one of the five solutions is a buffer solution.

(A)

Ethanoic acid CH₃COOH (a.k.a. acetic acid) is a weak acid. pKa = 4.756. CH₃COONa is a salt. It dissolves to produce CH₃COO⁻, which is the conjugate base ion of CH₃COOH. The solution in (A) contains equal number of CH₃COOH and CH₃COO⁻, both at 1.0 M.

Refer to the Henderson-Hasselbalch equation for buffers of weak acids.

$\displaystyle \text{pH} = \text{pK}_a + \log{\frac{[\text{Conjugate Ion}]}{[\text{Weak Acid}]}}$ .

$\displaystyle \log{\frac{[\text{Conjugate Ion}]}{[\text{Weak Acid}]}} =\ln{1} = 0$ .

The pH of the solution in (A) will be the same as the pKa of CH₃COOH. pH = 4.746.

(B)

Consider the hydrogen halides:

HF: weak acid.HCl: strong acid.HBr: strong acid.

The radius of halogen atoms increases down the group, and hydrogen-halogen bond becomes weaker. It becomes easier for water to break those bonds. As a result, the strength of hydrogen halides increases down the group. HF is the only weak acid among the common hydrogen halides.

Mixing HBr and KBr at equal ratio will be similar to mixing HCl and KCl at the same ratio. All HBr in the solution breaks down into H⁺ and Br⁻. The pH of the solution will depend only on the concentration of HBr.

$\displaystyle [\text{H}^{+}] = [\text{HBr}] \\\phantom{[\text{H}^{+}]}= \frac{n}{V} \\\phantom{[\text{H}^{+}]}= \frac{c(\text{HBr})\cdot V(\text{HBr})}{V(\text{HBr})+V(\text{KBr})}\\\phantom{[\text{H}^{+}]}=\frac{0.100\;\text{L}\times 1.0\;\text{mol}\cdot\text{L}^{-1}}{0.100\;\text{L}+0.100\;\text{L}} \\\phantom{[\text{H}^{+}]}= 0.50\;\text{mol}\cdot\text{L}^{-1}$ .

$\text{pH} = -\log{[\text{H}^{+}] = -\log{0.50} \approx {\bf 0.30}$ .

(C)

Similarly to HCl and HBr, HI is also a strong acid. Mixing HI and NaOH at equal ratio will produce a solution of NaI, which is similar to NaCl. The final solution will be neutral. pH = 7 if pKw = 14.

(D)

NH₃ is a weak base. NH₄Cl dissolves completely to produce NH₄⁺ and Cl⁻. NH₄⁺ is the conjugate acid of NH₃. The final solution will contain an equal number of NH₃ and NH₄⁺. pKb = 4.75 for ammonia NH₃.

Apply the Henderson-Hasselbalch equation for buffers of weak bases:

$\displaystyle \textbf{pOH} = \text{pK}_b + \log{\frac{[\text{Conjugate Ion}]}{[\text{Weak Base}]}}= 4.75 + \log{1} = 4.75$ .

Note that what this equation gives for buffers of weak bases is the pOH of the solution. pH = pKw - pOH. Assume that pKw = 14. pH = 14 - 4.75 = 9.25.

(E)

The solution in (E) will contain about 1.0 M of CH₃COOH. The volume of the solution will be 200 mL.

$n(\text{CH}_3\text{COO}^{-}) = n(\text{NaOH}] = c\cdot V = 0.10\;\text{mol}$ .

$\displaystyle [\text{CH}_3\text{COO}^{-}] = \frac{n}{V} = {0.10}{0.10 + 0.10} = 0.50 \;\text{mol}\cdot\text{L}^{-1}$ .

There's nearly no conjugate base of CH₃COOH. As a result, the solution will not be a buffer, and the Henderson-Hasselbalch Equation will not apply. Refer to an ICE table:

$\begin{array}{c|ccccccc}\text{R}&\text{CH}_3\text{COO}^{-} &+&\text{H}_2\text{O}&\rightleftharpoons &\text{CH}_3\text{COOH}&+&\text{OH}^{-}\\\text{I}&0.50\\\text{C}& -x &&&& +x &&+x\\\text{E} &0.50 - x &&&&x&&x\end{array}$

The value of pKa is large. Ka will be small. the value of will be much smaller than 0.50 such that $0.50-x \approx 0.50$ .

The pKa of a weak acid is the same as pKw divided by the pKb of its conjugate base.

$\displaystyle \frac{[\text{CH}_3\text{COOH}]\cdot[\text{OH}^{-}]}{[\text{CH}_3\text{COO}^{-}]} = \text{K}_b(\text{CH}_3\text{COO}^{-}) \\\phantom{\displaystyle \frac{[\text{CH}_3\text{COOH}]\cdot[\text{OH}^{-}]}{[\text{CH}_3\text{COO}^{-}]} }= \frac{\text{K}_w}{\text{K}_a(\text{CH}_3\text{COOH})} \\\phantom{\displaystyle \frac{[\text{CH}_3\text{COOH}]\cdot[\text{OH}^{-}]}{[\text{CH}_3\text{COO}^{-}]}} = \frac{10^{-14}}{1.75\times 10^{-5}} = 5.71\times 10^{-10}$ .

$\displaystyle \frac{x^{2}}{0.50} =5.71\times 10^{-10}$ .

$[\text{OH}^{-}] = x \approx 1.69\times 10^{-5}\;\text{mol}\cdot\text{L}^{-1}$ .

$\text{pH} = \text{pK}_w + \log{[\text{OH}^{-}]} = 9.23$ .

Answer from: Quest

the movement of aircraft through clouds can even cause lightning strikes. a single lightning event is a "flash", which is a complex, multi-stage explosively ejected.

explanation:

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Answer from: Quest

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is $1s^{2}2s^{2}2p^{6}3s^{1}$ and electronic configuration of phosphorous is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$ .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

$3na + p \rightarrow na_{3}p$

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.