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Juliette [100K]
2 years ago
9

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

hted in boldface is the acidic hydrogen atom? the ka value for lactic acid is 1.38 × 10−4.\?
Chemistry
1 answer:
Mashutka [201]2 years ago
6 0

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

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Dvinal [7]

Answer: All of the statements are true.

Explanation:

(a) Considering the system mentioned in the equation:-

The sum of total moles in the flask will always be equal to 1 which leads to confirmation of this statement as for 60 secs= 0.16 mol A and 0.84 mol B

(b) 0<t< 20s,  mole A got reduced from 1 mole to 0.54 moles while at 40s to 60s A got decreased from 0.30 moles to 0.16 moles.

0 to 20s is 0.46 (1 - 0.54 = 0.46)mol whereas,

40 to 60s is 0.14 (0.30-.16 = 0.14) mol

(0.46 > 0.14) mol leading this statement to be true as well.

(c) Average rate from t1 = 40 to t2 = 60 s is given by:

\delta moles/\delta time  = 0.30-0.16/60-40 = 0.007 Mol/s which is true as well

7 0
2 years ago
A) How many moles of CO2 and H2O are formed from 3.85 mole of propane C3H8 (This calculation needs to be done twice-once fro CO2
4vir4ik [10]

Answer:

a) 11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

15.4 moles of water are formed from 3.85 mole of propane.

b)0.5176 moles of water are formed from 0.647 mole of oxygen gas.

0.1294 moles of propane are consumed.

Explanation:

C_3H_7+5O_2\rightarrow 3CO_2+4H_2O

a) Moles of propane = 3.85 moles

According to reaction, 1 mole of propane gives 3 moles of carbon dioxide gas.

Then 3.85 moles of propane will give:

\frac{3}{1}\times 3.85 mol=11.55 mol of carbon dioxide gas.

11.55 moles of carbon dioxide gas are formed from 3.85 mole of propane.

According to reaction, 1 mole of propane gives 4 moles of water gas.

Then 3.85 moles of propane will give:

\frac{4}{1}\times 3.85 mol=15.4 mol of water .

15.4 moles of water are formed from 3.85 mole of propane.

b) Moles of oxygen gas = 0.647 moles

According to reaction, 5 mole of oxygen gas gives 4 moles of water.

Then 0.647 moles of oxygen will give:

\frac{4}{5}\times 0.647 mol=0.5176 mol of water.

0.5176 moles of water are formed from 0.647 mole of oxygen gas.

According to reaction, 5 mole of oxygen gas reacts with 1 mole of propane.

Then 0.647 moles of oxygen will give:

\frac{1}{5}\times 0.647 mol=0.1294 mol of propane.

0.1294 moles of propane are consumed.

5 0
2 years ago
How much energy is required to heat 0.24 KG lutetium from 296.2K to 373.5 K? The specific heat for lutetium is 0.154 J/g-K
Olenka [21]

I’m not sure I need help with this question

6 0
1 year ago
A white powder is added to a solution. The images show observations made before the powder is added, just after the powder has b
7nadin3 [17]

There was a change in its color from white to red which can only be changed by a chemical reaction

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At the beginning of the school year, a chalk company receives an order for 2000 boxes which is the largest order ever placed. Ea
lys-0071 [83]

Answer:

  • <u>259,000 g of chalk.</u>

Explanation:

<u>1) Data:</u>

a) 2000 boxes

b) 175 g / box

c) % yield = 74%

<u>2) Formula: </u>

  • % yield = (theoretical yield / actual yield) × 100

<u>3) Solution:</u>

a) Calcualte the actual yield:

  • mass of product = 2000 box × 175 g/ box = 350,000 g

b) Solve for the theoretical yield from the % yield formula:

  • % yield = (theoretical yield / actual yield) × 100

        ⇒ theoretical yield = % yield × actual yield / 100

             theoretical yield = 74% × 350,000g / 100 = 259,000 g

6 0
1 year ago
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