You mix 999 mL pure water and 1 mL of 2.0 M NaOH. Calculate the pH of the resulting solution. (assume the total volume is 1.0 L). a. 0.3
b. 0.7
c. 2.7
d. 11.3
e. 13.7

D => 11.3

Explanation:

(Molarity x Volume) diluted = (Molarity x Volume) concentrate

=> [NaOH] = (1/999)(2M) = 0.002M in NaOH = 0.002M OH⁻

pOH = -log[OH⁻] = -log(0.002) = 2.7

pH = 14 - pOH = 14 - 2.7 = 11.3

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 :