What is the pH of 10.0 mL solution of 0.75 M acetate after adding 5.0 mL of 0.10 M HCl (assume a Ka of acetic acid of 1.78x10-5)

The correct answer is 5.895.

Explanation:

The reaction will be,

CHCOO⁻ + H+ ⇔ CH₃COOH

Both the HCl and the acetate are having one n factor.

The millimoles of CH₃COO⁻ is,

= Volume in ml × molarity = 10 × 0.75 = 7.5

The millimoles of HCl = Volume in ml × molarity = 5 × 0.1 = 0.5

Therefore, 0.5 will be the millimoles of CH₃COOH formed, now the millimoles of the CH₃COO⁻ left will be, 7.5-0.5 = 7.0

The volume of the solution is, 10+5 = 15 ml

The molarity of CH₃COO⁻ is, millimoles / volume in ml = 7/15

The molarity of CH₃COOH is 0.5/15

pH = pKa + log[CH₃COO⁻]/[CH₃COOH]

= 4.74957 + 1.146

= 5.895

the correct option is c.

isotopes refers to different forms of the same element with the same atomic number but different mass number as a result of difference in their neutron numbers. isotopes have different relative atomic mass but share the same chemical properties.

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8