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Sidana [21]
3 years ago
10

How many of the following statements about silver acetate, AgCH3COO, are true? i) More AgCH3CoO(S) will dissolve if the pH of th

e solution is reduced to 2.5 i) Less AgCH3CO0(s) will dissolve if AgNO; (s) is added to the aqueous solution iii) More AgCH3CoO(s) will dissolve if NaOH (aq) is added to the aqueous solution a) 0 b) c) 2 d) 3
Chemistry
1 answer:
shutvik [7]3 years ago
8 0

Answer:

All three statements are true

Explanation:

Solubility equilibrium of silver acetate:

AgCH_{3}COO\rightleftharpoons Ag^{+}+CH_{3}COO^{-}

  • If pH is increased then concentration of H^{+} increases in solution resulting removal of CH_{3}COO^{-} by forming CH_{3}COOH. Hence, according to Le-chatelier principle, equilibrium will shift towards right. So more AgCH_{3}COO will dissolve
  • If AgNO_{3} is added then concentration of Ag^{+} increases in solution resulting shifting of equilibrium towards left in accordance with Le-chatelier principle. So less AgCH_{3}COO will dissolve
  • Insoluble precipitate of AgOH is formed by adding NaOH in solution resulting removal of Ag^{+}. So, more AgCH_{3}COO will dissolve

Hence all three statements are true

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A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa is:

pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

4 0
3 years ago
What is 70F converted into C
sweet [91]
It is 21.1 degrees C
5 0
3 years ago
Read 2 more answers
Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi
bija089 [108]

<u>Answer:</u> The value of K_c for the net reaction is \frac{K_b}{K_w}

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u>  B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b

<u>Equation 2:</u>  H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}

The net equation follows:

B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

K_c=K_1\times K_2

We are given:  

K_1=K_b

K_2=\frac{1}{K_w}

Putting values in above equation, we get:

K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}

Hence, the value of K_c for the net reaction is \frac{K_b}{K_w}

7 0
3 years ago
Complete the following single replacement reaction. If they don’t react, just write “NR”
stiv31 [10]

We have to complete all the given reactions.

1. Fe(s) + CuCl₂ → Cu + FeCl₂

2. Cu(s) + FeCl₂(aq)  → NR (no reaction takes place)

3. K(s) + NiBr2(aq) → NR (no reaction takes place)

4. Ni(s) + KBr(aq) → K + NiBr₂

5. Zn(s) + Ca(NO₃)₂(aq) → NR (no reaction)

6. Ca(s) + Zn(NO₃)₂(aq) → Zn(s) + Ca(NO₃)₂(aq)

4 0
3 years ago
If I have 2 liters of gas held at a pressure of 78 atm and a temperature of 900 k, what will be the volume of the gas if I decre
dangina [55]

Answer:

V2 = 2.88L

Explanation:

P1= 78atm, V1= 2L, T1= 900K, P2= 45atm, V2=? T2= 750K

Applying the general gas equation

P1V1/T1 = P2V2/T2

Substitute the above

(78*2)/900= (45*V2)/750

V2= (78*2×750)/(900*45)

V2= 2.88L

7 0
3 years ago
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