What is ph when 4.0 ml of 2.0 m barium hydroxide is added to 10.0 ml of 1.00 m nitric acid? 13.63 0.37 0.85 13.15

- pH sis 13.63

Solution:- It is a strong base vs strong acid titration. The equation for the reaction takes place between given acid and base is:

let's calculate the moles of each from given molarities and mL.

moles of barium hydroxide =

= 0.008 mol

moles of nitric acid =

= 0.01 mol

From balanced equation base and acid react in 1:2 mol ratio. So, let's calculate the moles of base used to react with the acid:

=

excess moles of barium hydroxide = 0.008 - 0.005 = 0.003 mol

Total volume of the solution = 0.004L + 0.010 mL = 0.014 L

Concentration of excess barium hydroxide =

= 0.214M

Barium hydroxide as two OH in it. So, the concentration of hydroxide ions will be twice of barium hydroxide concentration.

So,  = 0.428M

pOH = log(0.428)

pOH = 0.37

pH = 14 - pOH

pH = 14 - 0.37

pH = 13.63

First choice is correct, the pH of the solution is 13.63.

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.

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