Calculate the ph of a buffer prepared by mixing 500 ml of 0.38 m acetic acid with 500 ml of 0.25 m sodium acetate. pka for acetic acid is 4.76

(a) calculate the ph of a buffer prepared by mixing 400 ml of 0.24 m acetic acid with 300 ml of 0.48 m sodium acetate. (b) for the buffer in part a, calculate the new ph if to that buffer was added 10 ml of 6m naoh. ignore any volume change from the additional 30 ml.

pH = 4,76

a) pH = 4,94

b) pH = 5,51

Explanation:

To calculate the pH of a buffer you must use Henderson-Hasselbalch formula thus:

pH = pka + log

Where [A⁻] is acetate ion concentration and [HA] is acetic acid concentration

The concentration of acetic acid is:

0,5 L× 0,38 M ÷(0,5L+0,5L) =0,19 M

The concentration of acetate ion is:

0,5 L× 0,25 M ÷(0,5L+0,5L) = 0,125 M

The pH is:

pH = 4,76 + log

4,76

a) The concentration of acetic acid is:

0,4 L× 0,24 M ÷(0,4L+0,3L) =0,137 M

The concentration of acetate ion is:

0,3 L× 0,48 M ÷(0,4L+0,3L) = 0,206 M

The pH is:

pH = 4,76 + log

4,94

b) The reaction of NaOH with acetic acid produces acetate ions, thus, moles of NaOH will increase concentration of acetate ion and descrease acetic acid concentration, thus:

NaOH moles: 0,010L× 6M = 0,060 moles

The concentration of acetic acid is:

(0,4 L× 0,24 M) - 0,060 moles ÷(0,4L+0,3L) =0,0514 M

The concentration of acetate ion is:

(0,3 L× 0,48 M) + 0,060 moles ÷(0,4L+0,3L) = 0,291 M

The pH is:

pH = 4,76 + log

5,51

I hope it helps!

2b2t

explanation:

2b2t

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally