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3 years ago

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When 1.00 g of thiamine hydrochloride (also called vitamin B1 hydrochloride) is dissolved in water, then diluted to 10.00 mL, th

e pH of the resulting solution is 4.50. The formula weight of thiamine hydrochloride is 337.3 g/mol. Calculate Ka.

1 answer:

Semenov [28]3 years ago

4 0

Answer:

Ka= 3.37× $10^{-9}$

Explanation:

Ka is the measure of an acid's ability to donate H+ ions. A strong acid will have a larger Ka value. Consider the following equation

HA + H2O → H3O+ + A-

where,

HA is acid, which when dissolves in a solution dissociates into hydronium ion and A- ion. Ka will be represented as

Ka = $\frac{[H+] [A-]}{[HA]}$

Step 1

calculate the number of moles of thiamine hydrochloride

no of moles(n)=g/MM

n=1.00/337.3

n=2.96× $10^{-3}$

Step 2

The solution was diluted to 10.00ml

∵ 2.96× $10^{-3}$ / 0.01 L

=0.296M

Step 3

Find H+ using the formula

pH= -log [H+}

4.50= -log [H+]

[H]= $10^{-4.50}$

[H]=3.16× $10^{-5}$

Step 4

Substitude the above values in the formula

Ka= $\frac{[H+] [A-]}{[HA]}$

= $\frac{[0.0000316] [0.000316]}{0.296}$

= 3.37× $10^{-9}$

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2 years ago

At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T

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<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature of the solution = 298 K

Putting values in above equation, we get:

$12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M$

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

$1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

$0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (glucose) = 89.18 g

$M_{solute}$ = Molar mass of solute (glucose) = 180.16 g/mol

$W_{solvent}$ = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC$

Hence, the freezing point of solution is -0.974°C

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