Question

4. Forty-three milliliters of 0.130 M nitric acid is added to 30.0 mL of 0.095 M calcium hydroxide. Calculate the pH of the initial nitric acid solution and then calculate the pH of the combined solutions. Assume the solution volumes are additive. Hint: determine whether excess H+ or OH- remain in solution after neutralization, then calculate the pH.

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Answer 1

Given

[HNO3]= 0.130 M

Volume of HNO3 = 43 mL

Volume of Ca(OH)2 = 30.0 mL

[Ca(OH)2 = 0.095 M

Calculation of initial pH

Since nitric acid is strong acid and dissociates completely so the pH of the solution is directly calculated by using concentration of HNO3

[H3O+] = [HNO3]= 0.130 M

pH = - log [H3O+] = - log ( 0.130 )

= 0.886

Initial pH of the solution of nitric acid = 0.886

pH after addition of calcium hydroxide

mol of HNO3 = concentration x volume in L

= 0.130 M x 0.043 L

= 0.00559 mol

Mol of calcium hydroxide = 0.095 M x 0.030 L = 0.00285 mol

Lets show the reaction

2 HNO3 + Ca(OH)2 -- > Ca(NO3)2 +2 H2O

From this reaction we get mole ratio of HNO3 and calcium hydroxide

Calculation of limiting reactant

Moles of calcium hydroxide required to react with 0.00559 mol HNO3

= 0.00559 mol HNO3 x 1 mol calcium hydroxide / 2 mol HNO3

= 0.002795 mol calcium hydroxide

In actual there is 0.00285 mol of it so the calcium hydroxide is in excess

and HNO3 is limiting reactant which is consumed completely in this reaction

lets find out excess moles of calcium hydroxide

Excess moles of calcium hydroxide = original moels – reacted moles

= 0.00285 – 0.002795

= 5.5 E-5 mol calcium hydride

We know calcium hydroxide is strong base and its 1 mol gives 2 moles of OH- .

Moles of OH-

= 5.5 E-5 mol calcium hydroxide x 2 mol OH- / 1 mol calcium hydroxide

= 0.00011 mol OH-

[OH-]= mol OH- / total volume in L

= 0.00011 mol/ ( 0.043 + 0.030 ) L

= 0.0015 M

pOH = - log ([OH-]) = - log ( 0.0015 ) = 2.82

pH of the solution = 14 – pOH = 14-2.82= 11.18

pH of the solution = 11.18