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never [62]
3 years ago
5

Which of these compounds would you expect to be most soluble in water? ch3ch2ch2f ch3ch2ch2oh ch3ch2ch2ch3 ch3ch2cl description?

Chemistry
1 answer:
professor190 [17]3 years ago
3 0
Among the given compounds, <span>CH3CH2CH2OH (1-propanol) will have maximum solubility in water. 

Reason: 
Solubility is based on the principle of 'Like dissolves Like'. Hence, water is a polar solvent, hence it will dissolve polar solvent to maximum extent. Among the listed solutes, </span>CH3CH2CH2OH has maximum polarity. 
<span>
Further, </span>CH3CH2CH2OH is capable of displaying hydrogen bonding with water, while other compounds will not show hydrogen bonding. With reference to hydrogen bonding, it may be noted that when hydrogen atom is bonded to electronegative elements like O, N and F, it exhibits hydrogen bonding. Due to this, CH3CH2CH2OH will be readily soluble in water. <span>

</span>
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Calculate the poh of this solution. round to the nearest hundredth. ph = 1.90 poh =
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The answer is 5.10

Explanation:

<h3><u>Given</u>;</h3>
  • pH = 1.9
<h3><u>To </u><u>Find</u>;</h3>
  • pOH = ?

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pOH = 7 – pH

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3 years ago
determine the ph of a buffer that is 0.55 M HNO2 and 0.75 M KNO2. tha value of Ka for HNO2 is 6.8*10^-4
Mariana [72]

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

3 0
3 years ago
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