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3 years ago

What is the ratio of lactic acid (Ka = 1.37x^10-4) to lactate in a solution with pH =4.29

Chemistry

1 answer:

hram777 [196]3 years ago

6 0

Henderson–Hasselbalch equation is given as,

pH = pKa + log [A⁻] / [HA] -------- (1)

Solution:

Convert Ka into pKa,

pKa = -log Ka

pKa = -log 1.37 × 10⁻⁴

pKa = 3.863

Putting value of pKa and pH in eq.1,

4.29 = 3.863 + log [lactate] / [lactic acid]

Or,

log [lactate] / [lactic acid] = 4.29 - 3.863

log [lactate] / [lactic acid] = 0.427

Taking Anti log,

[lactate] / [lactic acid] = 2.673

Result:

2.673 M lactate salt when mixed with 1 M Lactic acid produces a buffer of pH = 4.29.

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When 100 mL of 1.0 M Na3PO4 is mixed with 100 mL of 1.0 M AgNO3, a yellow precipitate forms and [Ag ] becomes negligibly small.

lana [24]

Answer:

$Na3PO4 + 3AgNO3 -------> Ag3PO4 + 3NaNO3$

In which [Ag+] in negligibly small and the concentration of each reactant is 1.0 M

The answer is A) PO43- < NO3- < Na+

Explanation:

Ag+ is removed from the solution just like PO43-, so there are just 2 possible answers at this point: a or b. Then we can notice that Na3PO4 releases 3 moles of Na+ and just 1 mole of NO3-

We have 100mL of each reactant with the same concentration for both (1.0 M) so:

(0.1)(1)(3)= 0.3 mol Na+

(0.1)(1)= 0.1 mol NO3-

so PO43- < NO3- < Na+

5 0

2 years ago

What type of reaction is shown below c10H8+12O2 10co2+4H2O

MakcuM [25]

Answer:

Explanation:

Combustion. Have fun with that.

7 0

2 years ago

HELP ASAP

Bezzdna [24]

A. I think sorry if it’s wrong

5 0

2 years ago

Catalyst

Hoochie [10]

a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

$\tt \dfrac{4}{5}\times 0.5=0.4$

volume NO at 1273 K and 1 atm

$\tt V=\dfrac{nRT}{P}=\dfrac{0.4\times 0.08206\times 1273}{1}=41.785~L$

b. 15 L NH3 at STP ( 1mol = 22.4 L)

$\tt \dfrac{15}{22.4}=0.67~mol$

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

$\tt \dfrac{6}{4}\times 0.67=1$

mass H2O(MW = 18 g/mol) :

$\tt mass=mol\times MW=1\times 18=18~g$

c. mol NO at 1273 K and 1 atm :

$\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34$

mol ratio of NO : O2 = 4 : 5, so mol O2 :

$\tt \dfrac{5}{4}\times 0.34=0.425$

Volume O2 at STP :

$\tt 0.425\times 22.4=9.52~L$

5 0

2 years ago

Find the ratio of the effusion rate of hydrogen gas to the effusion rate of krypton gas.

Zielflug [23.3K]

This problem could be solved through the Graham’s law of effusion (also known as law of diffusion). This law states that the ratio of the effusion rate of the first gas and effusion rate of the second gas is equivalent to the square root of the ratio of its molar mass. Thus the answer would be 0.1098.

6 0

2 years ago

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