Chemistry, 16.11.2019 05:31 isabelle1670

The neutralization of h3po4 with koh is exothermic. h3po4(aq)+3koh(aq)⟶3h2o(l)+k3po4(aq )+173.2 kj if 60.0 ml of 0.200 m h3po4 is mixed with 60.0 ml of 0.600 m koh initially at 23.43 °c, predict the final temperature of the solution, assuming its density is 1.13 g/ml and its specific heat is 3.78 j/(g·°c). assume that the total volume is the sum of the individual volumes.

Answers: 3

Show answers

Answers

Answer from: makaylaspeaks4787

Final temperature of solution is 27.48 $^{0}\textrm{C}$

Explanation:

Total volume of mixture = (60.0+60.0) mL = 120.0 mL

We know, density = (mass)/(volume)

So mass of mixture = $(120.0\times 1.13)g=135.6 g$

Amount of heat released per mol of $H_{3}PO_{4}$ = $\frac{(m_{mixture}\times C_{mixture}\times \Delta T_{mixture})}{n_{H_{3}PO_{4}}}$

Where, m represents mass , C represents specific heat, $\Delta T$ represents change in temperature and n is number of moles

As this reaction is an exothermic reaction therefore temperature of mixture will be higher than it's initial temperature.

Let's say final temperature of mixture is T $^{0}\textrm{C}$

So, $\Delta T_{mixture}=(T-23.43)^{0}\textrm{C}$

Here $m_{mixture}=135.6 g$ and $C_{mixture}=3.78J/(g.^{0}\textrm{C})$

Moles of H_{3}PO_{4} are added = $\frac{0.200}{1000}\times 60.0moles$ = 0.012 moles

So, $(173.2\times 10^{3})J=\frac{[(135.6g)\times (3.78J.g^{-1}.^{0}\textrm{C}^{-1})\times (T-23.43)^{0}\textrm{C}]}{0.012}$

or, T = 27.48 $^{0}\textrm{C}$

So, final temperature of solution is 27.48 $^{0}\textrm{C}$

Answer from: Quest

consider the reaction below.

no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.