Calculate the ph of the following solutions (1) 0.02mhcl. (2) 0.10mnaoh. (3) 0.0032mhno. (4) 0.03mkoh

1. 1.70

2. 13.00

3. 2.49

4. 12.48

Explanation:

Background information and equations:

1. HCl is a strong acid, given its molarity:

2. NaOH is a strong base:

3. Nitric acid is a strong acid:

4. KOH is a strong base:

1) pH of HCl = 1.7

2) pH of NaOH = 13

3) pH of HNO₃ = 2.5

4) pH of KOH = 12.5

Explanation:

answer is in attachment

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.