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sergiy2304 [10]
3 years ago
12

20 mL of an approximately 10% aqueous solution of ethylamine, CH3CH2NH2, is titrated with 0.3000 M aqueous HCl. Which indicator

would be most suitable for this titration? The pKa of CH3CH2NH3 + is 10.75
Chemistry
1 answer:
Eduardwww [97]3 years ago
8 0

Answer:

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6

Bromocresol green, color change from pH = 4.0 to  5.6

Explanation:

The equation for the reaction is :

C_2H_5NH_2_(_a_q_)     +     H^+_(_a_q_)   ---      C_2H_5NH_{3(aq)}^+

concentration of C_2H_5NH_{2(aq) = 10%

10 g of C_2H_5NH_{2(aq) in 100 ml solution

molar mass = 45.08 g/mol

number of moles = 10 / 45.08

= 0.222 mol

Molarity of C_2H_5NH_2(aq) = 0.222 \times \frac{1000}{100}mL

= 2.22 M

number of moles of C_2H_5NH_{2(aq) in 20 mL can be determined as:

= 20 mL \times  2.22 M= 44*10^{-3} mole

Concentration of C_2H_5NH_2(aq) = \frac{44*10^{-3}*1000}{20}

= 2.22 M

Similarly, The pKa Value of C_2H_5NH_{2(aq) is given as 10.75

pKb value will be: 14 - pKa

= 14 - 10.75

= 3.25

the pH value at equivalence point is,

pH= \frac{1}{2}pKa - \frac{1}{2}pKb-\frac{1}{2}log[C]

pH = \frac{14}{2}-\frac{3.25}{2}-\frac{1}{2}log [2.22]

pH = 5.21

Therefore, The indicator that is best fit for the given titration is Bromocresol Green Color change from pH between 4.0 to 5.6

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d. IF3

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Need help ASAP!<br><br> How many moles of sodium nitrate are in 0.25 L of 1.2 M NaNO3 solution?
Setler79 [48]

Answer:

\boxed {\boxed {\sf 0.3 \ mol \ NaNO_3}}

Explanation:

Molarity is a measure of concentration in moles per liter.

molarity= \frac{moles \ of \ solute}{liters \ of \ solution}

The molarity of the solution is 1.2 M NaNO₃ or 1.2 moles NaNO₃ per liter. There are 0.25 liters of the solution. The moles of solute are unknown, so we can use x.

  • molarity= 1.2 mol NaNO₃/L
  • liters of solution=0.25 L
  • moles of solute =x

1.2 \ mol \ NaNO_3/L= \frac{x}{0.25 \ L}

We are solving for x, so we must isolate the variable, x. It is being divided by 0.25 liters. The inverse of division is multiplication, so we multiply both sides by 0.25 L.

0.25 \ L *1.2 \ mol \ NaNO_3/L=\frac{x}{0.25 \ L} *0.25 \ L

0.25 \ L *1.2 \ mol \ NaNO_3/L=x

The units of liters cancel, so we are left with the units moles of sodium nitrate.

0.25  *1.2 \ mol \ NaNO_3=x

0.3 \ mol \ NaNO_3=x

There are 0.3 moles of sodium nitrate.

3 0
2 years ago
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