What is the new pH?

Ten Examples

Every acid and every base in every example below is strong. Every single one.

Also, when these types of problems are discussed, the student will often just average the two pH values and think they have answered the problem correctly.

THEY WOULD BE WRONG.

Also, there are different scenarios possible:

(a) adding two acids together.

(b) adding an acid and a base together.

(c) adding two bases together.

Do not look at just Example #1 and conclude that all problems of mixing are solved like that.

In closing, a reminder:

You cannot just average the pH values of the two solutions and get a correct answer.

**Example #1:** Calculate the pH of a solution obtained by mixing 100. mL of an acid of pH = 3.00 and 400. mL solution of pH = 1.00.

**Solution:**

1) Determine total moles of H^{+} in solution:

for pH = 3.00, [H^{+}] = 0.0010 M

for pH = 1.00, [H^{+}] = 0.10 M(0.0010 mol/L) (0.100 L) = 0.00010 mol

(0.10 mol/L) (0.400 L) = 0.040 mol0.040 mol + 0.00010 mol = 0.0401 mol

Note: we ignore the H

^{+}contribution from water.

2) Determine the [H^{+}] of the combined solution:

0.0401 mol / 0.500 L = 0.0802 M

3) Determine the new pH:

pH = −log 0.0802 = 1.096To two sig figs, the answer is 1.10.

**Example #2:** Two solutions are mixed: 100. mL of HCl with a pH of 2.50 and 100. mL of NaOH with a pH of 11.00. What is the pH of the resulting solution?

**Solution:**

1) Determine [H^{+}] and [OH¯] present before any reaction:

[H^{+}] = 10¯^{pH}= 10¯^{2.50}= 0.0031623 M[OH¯] = 10¯

^{pOH}= 10¯^{3.00}= 0.00100 M

2) Determine millimoles of each reactant:

HCl ---> (0.0031623 mmol / mL) (100. mL) = 0.31623 mmolNaOH ---> (0.00100 mmol / mL) (100. mL) = 0.100 mmol

3) Allow them to react in a 1:1 molar ratio:

0.31623 mmol − 0.100 mmol = 0.21623 mmolIt should be clear that NaOH is the limiting reagent and will be 100% used up.

4) Determine the new hydrogen ion molarity:

0.21623 mmol / 200. mL = 0.00108115 M

5) Calculate the new pH:

pH = −log 0.00108115 = 2.97 (to two sig figs)

**Example #3:** Adding 5.0 mL of a solution at pH = 1.0 to 5.0 mL of a solution at pH = 7.0, what is my final pH?

**Solution:**

We will assume the pH = 7 solution is actually pure water (as opposed to a pH = 7 buffer).[H

^{+}] at pH = 1 ---> 10¯^{1}= 0.1 MThe dilution formula is M

_{1}V_{1}= M_{2}V_{2}:(0.1 mol/L) (5.0 mL) = (x) (10.0 mL)

x = 0.05 M

pH = −log [H

^{+}] = −log 0.05 = 1.3If the pH = 7 solution had been a buffer, the solution path would have been more complex and would require additional information about the buffer solution.

Note that the H

^{+}contribution for the pH = 7 water is ignored.

**Example #4:** Add 5.0 mL of a solution at pH = 6.0 to 5.0 mL of a solution at pH = 10.0. What is the final pH?

**Solution:**

[H^{+}] at pH = 6 ---> 10¯^{pH}= 1 x 10¯^{6}Mmoles H

^{+}in 5.0 mL ---> (1 x 10¯^{6}mol/L) (0.005 L) = 5 x 10¯^{9}mol[OH¯] at pH = 10 ---> 10¯

^{pOH}= 1 x 10¯^{4}Mmoles OH¯ in 5.0 mL ---> (1 x 10¯

^{4}mol/L) (0.005 L) = 5 x 10¯^{7}mol5 x 10¯

^{7}mol − 5 x 10¯^{9}mol = 4.95 x 10¯^{7}molThe step above treats the hydroxide as being in excess and reacting in a 1:1 molar ratio with all the acid present in the pH = 6.0 solution.

new [OH¯] ---> 4.95 x 10¯

^{7}mol / 0.010 L = 4.95 x 10¯^{5}Mnew pOH ---> −log 0.0000495 = 4.3

pH = 9.7

Comment: when the hydroxide reacted with the hydrogen ion, it converted the 5.0 mL of pH 6.0 solution to a solution of pH = 7.0, which is essentially pure water. This is why I did not react any more hydroxide. It simply became 0.0000495 moles of hydroxide ion in 10.0 mL of solution.

**Example #5:** Equal volumes of a pH = 4.0 and a pH = 6.0 solutions are mixed. What is the pH of the mixed solution that results?

The first solution is the more general solution, which MUST be used if the volumes are different. Solution #2 is a shorter calculation that works only if the volumes are equal.

**Solution #1:**

Let us assume 100. mL of each solution is mixed.

1) Moles of H^{+} in the pH = 4.0 solution:

pH = 4.0 means [H^{+}] = 0.0001 M(0.0001 mol/L) (0.100 L) = 0.00001 mol

2) Moles of H^{+} in the pH = 6.0 solution:

pH = 6.0 means [H^{+}] = 0.000001 M(0.000001 mol/L) (0.100 L) = 0.0000001 mol

3) Total moles of H^{+} in the mixed solution:

0.00001 mol + 0.0000001 mol = 0.0000101 mol

4) Calculate the new molarity:

0.0000101 mol / 0.200 L = 0.0000505 M

5) Calculate the pH:

pH = −log 0.0000505 = 4.3

**Solution #2:**

Since the volumes are the same, they drop out of consideration and the molarities can be used directly in the problem:

pH = 4 means [H^{+}] = 0.0001 M

pH = 6 means [H^{+}] = 0.000001 MThe sum of the molarities is 0.000101 M

Equal volumes means the solution doubles in volume.

Molarity of the combined solutions ---> 0.000101 / 2 = 0.0000505 M

pH = −log 0.0000505 = 4.3

**Example #6:** Consider the reaction of Ba(OH)_{2} and HCl. You are mixing 2.00 L of HCl solution that has pH of 1.52 and a solution of Ba(OH)_{2} that has pH = 13.30. What volume of the Ba(OH)_{2} solution is required to completely react with the HCl solution with no HCl or Ba(OH)_{2} remaining?

**Solution:**

1) Some information:

2HCl + Ba(OH)_{2}---> BaCl_{2}+ 2H_{2}OThe key molar ratio is the 2:1 ratio between HCl and Ba(OH)

_{2}. For every one mole of Ba(OH)_{2}that reacts, two moles of HCl are required.

2) Based on the pH of HCl (a strong acid, ionized 100%) equalling 1.52:

[H^{+}] = 10^{-1.52}= 0.0302 M

3) 2.00 L of 0.0302 M contains:

(0.0302 mol / L) (2.00 L) = 0.0604 mol of H^{+}

4) Based on the above equation:

Two mol HCl completely reacts with one mol Ba(OH)_{2}So 0.0604 mol HCl will require 0.0302 mol Ba(OH)

_{2}for complete reaction.

5) As for the Ba(OH)_{2}:

since pH = 13.30that means pOH = 14.00 − 13.30 = 0.70

[OH

^{-}] = 10^{-0.70}= 0.20 Mbecause Ba(OH)

_{2}---> Ba^{2+}+ 2OH^{-}[Ba(OH)

_{2}] = 0.10 M

6) Determine volume of 0.10 M Ba(OH)_{2} containing 0.0302 mol:

0.0302 mol / 0.10 mol L¯^{1}= 0.302 L = 302 mL (this is three sig figs)

7) However, two sig figs should be used for the answer. Therefore:

3.0 x 10^{2}mLNOT 300 mL because that is one sig fig.

NOT 300. mL because that is three sig figs.

**Example #7:** 20.20 mL of an HNO_{3} solution are needed to react completely with 300.0 mL of LiOH solution that has pH 12.05. What was the pH of the nitric acid solution?

**Solution:**

1) Molarity of the OH¯:

[OH¯] = 10¯^{pOH}= 10¯^{1.95}= 0.0112202 MRemember, LiOH is a strong base, so it ionizes 100%.

2) Moles of OH¯ present in LiOH solution

(0.0112202 mmol/mL) (300.0 mL) = 3.36606 mmol

3) H+^{} and OH¯ react in a 1:1 molar ratio, therefore:

3.36606 mmol of HNO_{3}was present in the 20.20 mL of solution

4) Molarity of the HNO_{3} solution:

3.36606 mmol / 20.20 mL = 0.1666366 M

5) pH of the nitric acid solution:

pH = −log 0.1666366 = 0.78

**Example #8:** Given 5.00 mL of 4.00 M NaOH mixed with 100. mL of HCl. After mixing, the pH is neutral. What was the pH of the HCl solution?

**Solution:**

1) Determine how many moles of NaOH are present:

(4.00 mmol/mL) (5.00 mL) = 20.0 mmolSince NaOH is a strong base, we have 20.0 mmol of OH¯ ion.

2) Since the pH of the mixed solution is neutral (pH = 7), we have no hydroxide ion (other than that present at pH = 7) in the solution. Since HCl and NaOH react in a 1:1 molar ratio, that means that 20.0 mmol of HCl was added.

3) What is the concentration of the HCl?

20.0 mmol / 100. mL = 0.200 MSince HCl is a strong acid, the [H

^{+}] = 0.200 M

4) Calculate the pH of the HCl solution:

pH = −log 0.200 = 0.699

**Example #9:** 25.00 mL of HNO_{3} solution with a pH of 2.12 is mixed with 25.00 mL of a KOH solution with a pH of 12.65. What is the pH of the final solution?

**Solution:**

1) Determine concentration of each reactant:

HNO_{3}---> 10¯^{2.12}= 0.0075858 M

KOH ---> 10¯^{1.35}= 0.0446684 M (note use of the pOH)

2) Determine millimoles of each reactant:

HNO_{3}---> (0.0075858 mmol/mL) (25.00 mL) = 0.189645 mmol

KOH ---> (0.0446684 mmol/mL) (25.00 mL) = 1.11671 mmol

3) The nitric acid is limiting. Use 1:1 molar ratio to determine KOH remaining:

1.11671 mmol − 0.189645 mmol = 0.927065 mmol

4) Determine new KOH molarity:

0.927065 mmol / 50.00 mL = 0.0185413 M

5) Determine pOH, then pH:

pOH = −log 0.0185413 = 1.73pH = 14 − 1.73 = 12.27

6) Because the volumes are equal, you can do this:

0.0446684 M − 0.0075858 M = 0.0370826 Mvolume doubles, so new molarity is 0.0185413 M

and so on to the new pH

If volumes are unequal, you must go the way outlined in steps 1 to 5.

**Example #10:** If solution A has a pH of 4.3, and solution B has a pH of 6.5 what is the pH after equal volumes are mixed?

**Solution:**

To the student: there may be a big temptation on your part to simply average 4.3 and 6.5 and think you have answered the question correctly. You would be wrong.

1) Determine the [H^{+}] for each solution:

A ---> [H^{+}] = 10¯^{4.3}= 0.0000501187 M

B ---> [H^{+}] = 10¯^{6.5}= 0.000000316228 M

2) Determine millimoles (assume 1.00 mL of each solution is present):

A ---> (0.0000501187 mmol/mL) (1.00 mL) = 0.0000501187 mmol

B ---> (0.000000316228 mmol/mL) (1.00 mL) = 0.000000316228 mmol

3) Add millimoles and determine new molarity of combined solutions:

0.0000501187 mmol + 0.000000316228 mmol = 0.000050434928 mmol0.000050434928 mmol / 2.00 mL = 0.000025217464 M

4) Determine pH:

pH = −log 0.000025217464 = 4.6

5) Because the volumes are equal, you can do this:

0.0000501187 M + 0.000000316228 M = 0.000050434928 MBecause the two molarities are diluted by half (remember equal volumes are added):

0.000050434928 M / 2 = 0.000025217464 M

And from there to the pH.

This only works with equal volume. With unequal volumes, you must determine millimoles, then add then, then determine new molarity.

**Example #11:** Two acidic solutions are to be mixed together. Solution "A" has a pH of 5.90 and solution "B" has a pH of 1.80. If 2.0 liters of solution "A" are mixed with 1.3 liters of solution "B", calculate the pH of the resulting solution.

**Solution:**

1) Use this equation first:

[H^{+}] = 10¯^{pH}[H

^{+}] of solution A ---> 10¯^{5.90}= 0.000001259 M

[H^{+}] of solution B ---> 10¯^{1.80}= 0.0158489 M

2) Determine moles of [H^{+}] in each solution:

solution A ---> (0.000001259 mol/L) (2.0 L) = 0.000002518 mol

solution B ---> (0.0158489 mol/L) (1.3 L) = 0.02060357 mol

3) Add moles:

0.000002518 mol + 0.02060357 mol = 0.020606088 mol

4) Determine molarity of mixed solutions:

0.020606088 mol / 3.3 L = 0.00624427 M <--- new conc of hydrogen ion

5) Determine pH of mixed solution:

pH = −log 0.00624427 = 2.20

**Bonus Example:** There are two liquids, A and B. A has a pH of around 4.80. Liquid B cannot drop below a pH of about 5.60 before being ruined, and it has a pH of 6.50 itself. How much of A can be added in with B, without B being ruined (in ratio form)?

**Solution:**

1) Convert the pH values to concentrations:

[H^{+}] in A ---> 10¯^{4.80}= 1.585 x 10¯^{5}M[H

^{+}] in B initially ---> 10¯^{6.50}= 3.162 x 10¯^{7}M[H

^{+}] in B at the edge of being ruined ---> 10¯^{5.60}= 2.512 x 10¯^{6}M

2) Let V_{A} be the volume of A to be added, and V_{B} be the volume of B to begin with. (In the end, the ratio asked for will be the value of V_{A} / V_{B}.)

3) Supposing additive volumes:

[(V_{A}) (1.585 x 10¯^{5}M) + (V_{B}) (3.162 x 10¯^{7}M)] / (V_{A}+ V_{B}) = 2.512 x 10¯^{6}M

4) Solve for V_{A}/V_{B}. First step: multiplying both sides by 10^{7}:

[(V_{A}) (158.5) + (V_{B}) (3.162)] / (V_{A}+ V_{B}) = 25.12(158.5 V

_{A}) + (3.162 V_{B}) = 25.12 (V_{A}+ V_{B})158.5 V

_{A}+ 3.162 V_{B}= 25.12 V_{A}+ 25.12 V_{B}133.38 V

_{A}= 21.958 V_{B}V

_{A}/ V_{B}= 21.958 / 133.38 = 0.1646Or, roughly 6.1 parts A to 1 part B by volume.