Determine the pH during the titration of 28.5 mL of 0.102 M hydrochloric acid by 0.272 M sodium hydroxide at the following points:

Question

Determine the pH during the titration of 28.5 mL of 0.102 M hydrochloric acid by 0.272 M sodium hydroxide at the following points:

(1) Before the addition of any sodium hydroxide

(2) After the addition of 5.35 mL of sodium hydroxide

(3) At the equivalence point 

(4) After adding 13.2 mL of sodium hydroxide

Answer 1

First, let's take a look at the acid-base reaction. The equation for this one is:

HCl + NaOH --> H2O + Cl- + Na+

We also need to know the equation for pH, which is:

pH = -log[H+].

(1) Before the addition of any sodium hydroxide, you know that you have a solution that is 0.102M [HCl]

Since [H] = [Cl-] = [HCl], you know that [H+] = 0.102.

1. pH=-log(0.102) = 0.991

(2) Now, you need to know how many moles of your acid were neutralized by the NaOH.

1. 5.35 mL NaOH * (1L/1000mL) * (0.272mol NaOH/L) = 0.0015 moles NaOH

To find out how many moles of HCl you started with:

1. 28.5mL HCl * (1/1000mL) * (0.102mol HCl/L) = 0.0029 moles HCl

So, since you started with 0.0029 moles of HCl and you added 0.0015 moles of NaOH, you

have 0.0014 moles of HCl leftover. (see the equation above)

To find the molarity, you will take the

total volume of your solution, which is 33.85 mL (Initial 28.5 mL + 5.35 mL).Now, remember that

molarity is moles/L, so you will need to convert 33.85 mL into L to get your molarity.

You should obtain a final concentration of 0.0414M.

1. pH = -log(0.0414) = 1.38
2. This pH makes sense because you are adding a base to your solution and therefore you are making it more basic

(3) At the equivalence point, the number of moles of acid and base are equal to each other. In this case,

that means you should have neutralized all of the acid with your base. For practice, you should figure out the volume of NaOH needed for this. Since everything is equal,

1. [OH-] = [H+] and pH = pOH
2. Therefore, since pH + pOH = 14, the pH is equal to 7 and is neutral

(4) First, you will need to find the number of moles of NaOH, just as you did in (2),

1. 13.2mL NaOH * (1L/1000mL) * (0.272 moles NaOH/L) = 0.0036 moles NaOH.

Since we discovered in (2) that the starting moles of HCl was 0.0029, we don't need to do that

calculation again.

We know that the 0.0029 moles of HCl is neutralized by 0.029 moles of NaOH, which leaves 0.0007

moles of NaOH remaining.

1. The total volume of the solution at this point is 28.5mL + 13.2 mL = 41.7 mL = 0.0417L.
2. The [OH] (concentration of OH in moles/L) is therefore 0.0007moles/0.0417L = 0.0168M
3. Just as we previously did for the relation between pH and [H+], we can use this formula:
4. pOH = -log[OH-]
5. Therefore, pOH = -log(0.0168) = 1.78

In order to find pH, we must remember that pH + pOH = 14. Substituting our value of pOH into this

equation, we find that pH = 12.22

Answer 2

Assumptions:

1. NaOH and HCl are strong base and acid respectively. This means that the reaction will go to completion until all of the limiting reactant is consumed.
2. Volumes of solutions are additive.
3. pH = 0.99 HCl is a strong so the concentration of H⁺ is equal to the molarity of the HCL solution.
4. pH = 1.37 (NaOH is limiting)
5. pH = 7.00 Neither reactant is limiting and the since both acid and base are strong, the resulting solution of NaCl is neither acid or basic.
6. pH = 12.21 (HCl is limiting)