What is the concentration, in M, of a KBr solution that contains 0.0688 moles of

A beaker with 1.60×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and c

stich3 [128]

Answer:

The pH will change 0.16 ( from 5.00 to 4.84)

Explanation:

Step 1: Data given

volume of acetic acid buffer = 160 mL

The total molarity of acid and conjugate base in this buffer is 0.100 M

A student adds 7.10 mL of a 0.460 M HCl solution to the beaker.

The pKa of acetic acid is 4.740

pH = 5.00

Step 2: Calculate concentration of acid

Consider x = concentration acid

Consider y = concentration conjugate base

x + y = 0.100

5.00 = 4.740 + log y/x

5.00 - 4.740 = log y/x

0.26 = log y/x

10^0.26 =1.82 = y/x

1.82 x = y

Since x+y = 0.100

x + 1.82 x = 0.100

2.82 x = 0.100

x =0.0355 M = concentration acid

Step 3: Calculate concentration of conjugate base

y = 0.100 - x

0.100 - 0.0355 =0.0645 M= concentration conjugate base

Step 4: Calculate moles of acid

Moles = volume * molarity

moles acid = 0.160 L * 0.0355 M= 0.00568 moles

Step 5: Calculate moles of conjugate base

moles conjugate base = 0.0645 M * 0.160 L=0.01032 moles

Step 6: Calculate moles HCl

moles HCl = 7.10 * 10^-3 L * 0.460 M=0.003266 moles

Step 7: Calculate new moles

A- + H+ = HA

moles conjugate base = 0.01032 - 0.003266 =0.007054 moles

moles acid = 0.00568 + 0.003266=0.008946 moles

Step 8: Calculate the total volume

total volume = 160 + 7.10 = 167.1 mL = 0.1671 L

Step 9: Calculate the concentration of the acid

concentration acid = 0.008946/ 0.1671 =0.0535 M

Step 10: Calculate the concentration of conjugate base

concentration conjugate base = 0.007054/ 0.1671 =0.0422 M

Step 11: Calculate the pH

pH = 4.740 + log 0.0535/ 0.0422=4.84

change pH = 5.00 - 4.84=0.16

The pH will change 0.16

5 0

3 years ago

Write the molecular formula for a compound with the possible elements C, H, N and O that exhibits a molecular ion at M

liberstina [14]

Answer:

$= \mathbf{C_3H_6O}$

Explanation:

From the given information, since the molecular mass of the ion M+ is not given;

Let's assume M+ = 58.0423

So, by applying the 13th rule;

we will need to divide the mass by 13, after dividing it;

The quotient n = no. of carbon; &

The addition of the quotient (n) with the remainder r = no. of hydrogen.

So;

$\dfrac{58}{13}= 4 \ remainder \ 6$

So;

$C_nH_{n+r} = C_4H_{4+6}$

$= C_4H_{10}$

From the given information; we have oxygen present, so since the mass of oxygen = 16, we put oxygen in the molecular formula by removing $CH_4$ . Also, since the mass is an even number then Nitrogen is 0.

So, we have:

$= \mathbf{C_3H_6O}$

6 0

2 years ago

There are two naturally occurring isotopes of copper. 63cu has a mass of 62.9296 amu. 65cu has a mass of 64.9278 amu. determine

SSSSS [86.1K]

1) You need to use the atomic mass of copper.

You can find it in a periodic table. It is 63.546 amu.

2) The atomic mass is the weigthed mass of the different isotopes.

This is, the atomic mass of one element is the atomic mass of each isotope times its corresponding abundance:

=> atomic mass of the element = abundance isotope 1 * atomic mass isotope 1 + abundance isotope 2 * atomic mass isotope 2 + ....+abundance isotope n * atomic mass isotope n.

3) The statement tells there are two isotopes so the abundance of one is x and the abundance of the other is 1 - x

=> 63.546 amu = x * 62.9296 amu + (1-x)*64.9278

=> 63.546 = 62.9296x + 64.9278 - 64.9278x

=> 64.9278x - 62.9296 = 64.9278 - 63.546

=> 1.9982x = 1.3818

=> x = 1.3818 / 1.9982 = 0.6915 = 69.15%

=> 1 - x = 1 - 0.6915 = 0.3085 = 30.85%

Answer:

Cu-63 69.15%;

Cu-65 : 30.85%

3 0

2 years ago

________ exercise strengthens skeletal muscle, and ________ exercise strengthens skeletal and cardiac muscle.

valentinak56 [21]

Answer:

6 0

2 years ago

Why it is difficult to accurately represent ammonia two-dimensionally?

Svetlanka [38]

According to the valence shell electron pair repulsion (VSPER) theory, an ammonia molecule <span> has a </span>trigonal pyramidal<span> shape with an experimental bond angle measure of 106.7 degrees. This is why it is difficult to accurately represent ammonia two-dimensionally because the molecular structure entails a 3-D projection with angles in it unlike the linear structure.</span>

8 0

2 years ago