Answer:
pH = 3.3
Explanation:
Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.
In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].
Solution using the I.C.E. table:
HNO₂ ⇄ H⁺ + KNO₂⁻
C(i) 0.55M 0M 0.75M
ΔC -x +x +x
C(eq) 0.55M - x x 0.75M + x b/c [HNO₂] / Ka > 100, the x can be
dropped giving ...
≅0.55M x ≅0.75M
Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]
=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M
pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3
Solution using the Henderson-Hasselbalch Equation:
pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]
= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]
= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3
Answer:As predicted by VSEPR theory, the molecule adopts a "bent" molecular geometry similar to that of water.
Explanation:
Answer:
Once an enzymatic reaction is completed, the enzyme releases substrates.
Explanation:
The enzyme will always return to its original state at the completion of the reaction. One of the important properties of enzymes is that they remain ultimately unchanged by the reactions they catalyze. After an enzyme is done catalyzing a reaction, it releases its products (substrates).
Answer:
M
Explanation:
Concentration of ![Cu^{2+} = [Cu(NO_3)_2]](https://tex.z-dn.net/?f=Cu%5E%7B2%2B%7D%20%3D%20%5BCu%28NO_3%29_2%5D)
= 0.020 M
Constructing an ICE table;we have:
![Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}](https://tex.z-dn.net/?f=Cu%5E%7B2%2B%7D%2B4NH_3_%7Baq%7D%20%5Crightleftharpoons%20%5BCu%28NH_3%29_4%5D%5E%7B2%2B%7D_%7B%28aq%29%7D)
Initial (M) 0.020 0.40 0
Change (M) - x - 4 x x
Equilibrium (M) 0.020 -x 0.40 - 4 x x
Given that: 
![K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}](https://tex.z-dn.net/?f=K_f%20%7D%20%3D%20%5Cfrac%7B%5BCu%28NH_3%29_4%5D%5E%7B2%2B%7D%7D%7B%5BCu%5E%7B2%2B%7D%5D%5BNH_3%5D%5E4%7D)
Since x is so small; 0.40 -4x = 0.40
Then:
M
