Given a diprotic acid, H2A, with two ionization constants of Ka1=2.3×10^− 4 and Ka2=3.8×10^−12, calculate the pH for a 0.142M solution of NaHA.

The pH for a 0.142M solution of NaHA is  1.0

Explanation:

Calculating the PH value based on the given data H2A

H2A with Ka1 = 2.3× 10–4

The concentration of the acid is 0.180 M and the Ka1 is 2.3×10–4

Rewrite the chemical equation as,

H2A --> H+1 + HA−1 with Ka1 = 2.3× 10–4

Bythe formula:

Ka=(products) / (reactants):

2.3× 10–4 = (x2)/(0.180)

x2= 0.180 × 2.3× 10–4

x= 0.414 ×10^-4 =[H+]

x= 4.14 ×10^-3 =[H+]

Since pH = -log of hydrogen ion concentration,

The pH for a 0.142M solution of NaHA is  1.0

The structure of indium consists of positive ions/cations and a sea of delocalised electrons attracting each other to form strong metallic bonds. Explanation: ... The metallic bond is the electrostatic forces of attraction between the delocalised electrons and positive ions.