part 2: for each of the problems below, assume 100% dissociation.\n1. a. write the equation for the…

1.

Answer:

A.

$HCl(aq)\rightarrow H^{+}(aq)+Cl^{-}(aq)$

B.

Answer:

2.32

Explanation:

Step1: Identify $[H^{+}]$ concentration

Since $HCl$ is a strong - acid and dissociates completely, $[H^{+}]=0.00476\ M$

Step2: Calculate pH

$pH =-\log[H^{+}]=-\log(0.00476)\approx2.32$

2.

Answer:

A.

$H_{2}SO_{4}(aq)\rightarrow 2H^{+}(aq)+SO_{4}^{2 -}(aq)$

B.

Answer:

1.34

Explanation:

Step1: Calculate molarity of $H_{2}SO_{4}$

The molar mass of $H_{2}SO_{4}$ is $M=(2\times1 + 32+4\times16)=98\ g/mol$. The number of moles of $H_{2}SO_{4}$, $n=\frac{m}{M}=\frac{3.25\ g}{98\ g/mol}\approx0.0332\ mol$. The molarity $M=\frac{n}{V}=\frac{0.0332\ mol}{2.75\ L}\approx0.0121\ M$.

Step2: Determine $[H^{+}]$ concentration

Since $H_{2}SO_{4}$ is a strong - acid and gives 2 moles of $H^{+}$ per mole of $H_{2}SO_{4}$, $[H^{+}]=2\times0.0121\ M = 0.0242\ M$

Step3: Calculate pH

$pH=-\log[H^{+}]=-\log(0.0242)\approx1.34$

3.

Answer:

A.

$NaOH(aq)\rightarrow Na^{+}(aq)+OH^{-}(aq)$

B.

Answer:

10.92

Explanation:

Step1: Identify $[OH^{-}]$ concentration

Since $NaOH$ is a strong - base and dissociates completely, $[OH^{-}]=0.000841\ M$

Step2: Calculate pOH

$pOH =-\log[OH^{-}]=-\log(0.000841)\approx3.08$

Step3: Calculate pH

$pH = 14 - pOH=14 - 3.08 = 10.92$

4.

Answer:

A.

$Al(OH)_{3}(s)\rightleftharpoons Al^{3 +}(aq)+3OH^{-}(aq)$

B.

Answer:

$7.08\times10^{-5}\ M$

Explanation:

Step1: Calculate pOH

$pOH=14 - pH=14 - 9.85 = 4.15$

Step2: Calculate $[OH^{-}]$ concentration

$[OH^{-}]=10^{-pOH}=10^{-4.15}\approx7.08\times10^{-5}\ M$

Step3: Determine $[Al(OH)_{3}]$ concentration

From the dissociation equation, for every mole of $Al(OH){3}$ that dissociates, 3 moles of $OH^{-}$ are produced. So $[Al(OH){3}]=\frac{[OH^{-}]}{3}=\frac{7.08\times10^{-5}\ M}{3}\approx2.36\times10^{-5}\ M$ (assuming complete dissociation and neglecting the solubility - product considerations for simplicity as per the 100% dissociation assumption). But if we consider the stoichiometry directly from the dissociation, since $[OH^{-}]$ is related to $[Al(OH){3}]$ by a 3:1 ratio, and we know $[OH^{-}]$, the concentration of $Al(OH){3}$ that would produce this $[OH^{-}]$ is $\frac{[OH^{-}]}{3}$. Since $[OH^{-}]=7.08\times10^{-5}\ M$, the concentration of $Al(OH)_{3}$ is $7.08\times10^{-5}\ M$ (because we are working backward from the $[OH^{-}]$ value to the original base concentration based on the stoichiometry of the dissociation).

5.

Answer:

A.

$Ca(OH)_{2}(s)\rightarrow Ca^{2 +}(aq)+2OH^{-}(aq)$

B.

Answer:

0.117 g

Explanation:

Step1: Calculate pOH

$pOH=14 - pH=14 - 11.64 = 2.36$

Step2: Calculate $[OH^{-}]$ concentration

$[OH^{-}]=10^{-pOH}=10^{-2.36}\approx4.37\times10^{-3}\ M$

Step3: Calculate moles of $OH^{-}$

The volume of the solution $V = 2.55\ L$. The number of moles of $OH^{-}$, $n_{OH^{-}}=[OH^{-}]\times V=4.37\times10^{-3}\ M\times2.55\ L\approx0.0111\ mol$

Step4: Calculate moles of $Ca(OH)_{2}$

From the dissociation equation, 1 mole of $Ca(OH){2}$ gives 2 moles of $OH^{-}$. So the number of moles of $Ca(OH){2}$, $n_{Ca(OH){2}}=\frac{n{OH^{-}}}{2}=\frac{0.0111\ mol}{2}=0.00555\ mol$

Step5: Calculate mass of $Ca(OH)_{2}$

The molar mass of $Ca(OH){2}$ is $M=(40+(2\times(16 + 1)))=74\ g/mol$. The mass of $Ca(OH){2}$, $m=n_{Ca(OH)_{2}}\times M=0.00555\ mol\times74\ g/mol = 0.117\ g$