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UkoKoshka [18]
2 years ago
8

The pH of a 0.25 M aqueous solution of hydrofluoric acid, HF, at 25.0 °C is 2.03. What is the value of Ka for HF?

Chemistry
2 answers:
Mazyrski [523]2 years ago
8 0
Last option that is none of above is right answer.
MrRa [10]2 years ago
7 0

Answer : The correct option is, 3.5\times 10^{-4}

Solution : Given,

pH = 2.03

Concentration of HF = 0.25 M

First we have to calculate the concentration of H^+ ion.

pH=-\log [H^+]

2.03=-\log [H^+]

[H^+]=9.3\times 10^{-3}M

Now we have to calculate the value of K_a for HF.

The equilibrium reaction will be

HF\rightleftharpoons H^++F^-

Concentration of H^+ = Concentration of F^- = 9.3\times 10^{-3}M

The expression for K_a for HF will be,

K_a=\frac{[H^+][F^-]}{[HF]}=\frac{(9.3\times 10^{-3})\times (9.3\times 10^{-3})}{0.25}=3.5\times 10^{-4}

Therefore, the value of K_a for HF is, 3.5\times 10^{-4}

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For the following equilibrium: A+2B⇋C+3D If the change in concentration for B is 0.44 M, what will be the change in concentratio
Readme [11.4K]

Answer:0.22M

Explanation:

5 0
2 years ago
Consider the standard galvanic cell based on the following half-reactions The electrodes in this cell are and . Does the cell po
Free_Kalibri [48]

Question: The question is incomplete and can't be comprehended. See the complete question below and the answer.

Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80 V Cu2+ + 2 e- → Cu 0.34 V

How would the following changes alter the potential of the cell?

a) Adding Cu2+ ions to the copper half reaction (assuming no volume change).

b) Adding equal amounts of water to both half reactions.

c) Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

d) Adding Ag+ ions to the silver half reaction (assume no volume change)

Explanation:

Nernst equation relates the reduction potential of an electrochemical reaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation.

Reaction under consideration:

Ag+ + e- → Ag 0.80 V

Cu+2 + 2 e- → Cu 0.34 V

Clearly, Ag reduction potential is high and this indicates that it gets reduced readily which leaves Cu to oxidize. Cu+2 ions are products of reaction and Ag+ ions are reactant ions.

Nernst equation : Ecell = E°cell­ – (2.303 RT / n F) log Q    

where                            

             Ecell = actual cell potential

             E°cell­ ­​ = standard cell potential

             R = the universal gas constant = 8.314472(15) J K−1 mol−1

             T = the temperature in kelvins

              n = the number of moles of electrons transferred                                    

 F = the Faraday constant, the number of coulombs per mole of electrons:

  (F = 9.64853399(24)×104 C mol−1)

 Q = [product ion]y / [reactant ion] x

Accordingly when applied to above reaction one will get the following

= E°cell­ – (2.303 x RT / 6 F) log [Cu+2] / [Ag+]

Now the given variables can be studied according to Le Chatelier's principle which states when any system at equilibrium is subjected to change in its concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.

a)        Adding Cu2+ ions to the copper half reaction (assuming no volume change).

Addition of Cu+2 ions increases its concentration and consequently increases the Q value which results in reduction of Ecell. In other words the addition of Cu+2 ions favors the backward reaction to maintain the equilibrium of reaction and hence the forward reaction rate decreases.

b)       Adding equal amounts of water to both half reactions.

Addition of water increases the dilution of the electrochemical cell. For weak electrolytes such as Ag+/ Cu+2 with increase in dilution, the degree of dissociation increases and as a result molar conductance increases.

c)        Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

Based on Le Chatelier's principle when Cu+2 ions amount is decreased by its continuous removal from  the system the forward reaction is favored. As the Cu+2 ions is removed the system attempts to generate more Cu+2 ions to counter the affect of its removal.

d)       Adding Ag+ ions to the silver half reaction (assume no volume change)

Addition of reactant ions, i.e. Ag+ ions, will favour the forward reaction, which results in more product formation.

6 0
2 years ago
What is the name of the following ionic compound?: MoAs
Minchanka [31]
Molybdenum Arsenide

I think that’s right but not %100 sure
3 0
2 years ago
Iron (Fe) undergoes an allotropic transformation at 912°C: upon heating from a BCC (α phase) to an FCC (γ phase). Accompanying t
-Dominant- [34]

Answer:

The description including its given problem is outlined in the following section on the clarification.

Explanation:

The given values are:

RBCC = 0.12584 nm

RFCC = 0.12894 nm

The unit cell edge length (ABCC) as well as the atomic radius (RBcc) respectively connected as measures for BCC (α-phase) structure:

√3 ABCC = 4RBCC

⇒  ABCC = \frac{4RBCC}{\sqrt{3} }

⇒             = \frac{4\times 0.12584}{\sqrt{3}}

⇒             = 0.29062 \ nm

Likewise AFCC as well as RFCC are interconnected by  

√2AFCC = 4RFCC

⇒  AFCC = \frac{4RFCC}{\sqrt{2}}

⇒             = \frac{4\times 0.12894}{\sqrt{2} }

⇒             = 0.36470 \ nm

Now,

The Change in Percent Volume,

= \frac{V \ final-V \ initial}{V \ initial}\times 100 \ percent

= \frac{(VFCC)unit \ cell-(VBCC)unit \ cell}{(VBCC)unit \ cell}\times 100 \ percent

= \frac{(aFCC)^3-(aBCC)^3}{(aBCC)^3}\times 100 \ percent

= \frac{(0.36470)^3-(0.29062)^3}{(0.29062)^3}\times 100 \ percent

= 97.62 \ percent (approximately)

Note: percent = %

7 0
2 years ago
Which of the following statements is true of an aqueous solution of sodium chloride? A]One that has 30.0 grams of NaCl dissolved
andrew11 [14]
<span>One that has 30.0 grams of NaCl dissolved in 100 grams of water is more concentrated than one that has 15.0 grams of NaCl dissolved in 100 grams of water. A is the correct answer.</span>
3 0
2 years ago
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