Ph is a logarithmic scale used to indicate the hydrogen ion concentration, [h+], of a solution: ph=−log[h+] due to the autoionization of water, in any aqueous solution, the hydrogen ion concentration and the hydroxide ion concentration, [oh−], are related to each other by the kw of water: kw=[h+][oh−]=1.00×10^−14 where 1.00×10^−14 is the value at approximately 297 k. based on this relation, the ph and poh are also related to each other as 14.00=ph+poh. the temperature for each solution is carried out at approximately 297 k where kw=1.00×10^−14. part a) 0.35 g of hydrogen chloride (hcl) is dissolved in water to make 2.5 l of solution. what is the ph of the resulting hydrochloric acid solution? express the ph numerically to two decimal places.

part b) 0.80 g of sodium hydroxide (naoh) pellets are dissolved in water to make 2.0 l of solution. what is the ph of this solution?
express the ph numerically to two decimal places.

Explanation:

The dissolution of HCl is  HCl → H⁺ + Cl⁻

The problem gives us V=2.5 L, and the moles can be calculated using the molecular weight of HCl, 36.46 g/mol:

= 9.60*10⁻³ mol H⁺

So the concentration of H⁺ is

[H⁺] = 9.60*10⁻³ mol / 2.5 L = 3.84 * 10⁻³ M

pH = -log [H⁺] = -log (3.84 * 10⁻³) = 2.42

= 0.02 mol OH⁻

[OH⁻] = 0.02 mol / 2.0 L = 0.01

pOH = -log [OH⁻] = -log (0.01) = 2.00

With the pOH, we can calculate the pH:

pH + pOH = 14.00

pH + 2.00 = 14.00

pH = 12.00

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

the correct answer would be d. hope this !