# CALCULATIONS FOR STRONG ACIDS AND BASES

Examples 1.21 to 1.28 represent routinely encountered problems and calculations for strong acids and bases. In general, 1 mole of acid (H+ ion) neutralizes 1 mole of base (OH- ion) to produce 1 mole of water (H2O).

Example 1.21

An aqueous solution of hydrochloric acid is 0.034 M. Find:

A. [H+]

B. [Cl-]

Solution

Since HCl is a strong acid, it dissociates and ionizes completely (100%) into its component ions. Thus,

A. [H+] = 0.034 M

B. [Cl-] = 0.034 M

It is safe to assume that [HCl] in terms of molecules is 0.

Example 1.22

For a 0.035 M HNO3 solution, find:

A. [H+]

B. pH

Solution

A. HNO3 is a strong acid. Thus, a 0.035 M solution of HNO3 yields 0.035 M H+ ions (and 0.035 M NO3- ions), since there is a complete dissociation and ionization.

[H+] = 0.035 M

B. pH = -log[H+] = -log(0.035 M) = 1.46 Example 1.23

Find [OH-] for the solution in Example 1.22.

Solution

Since this is an aqueous solution, the ion-product constant expression Kw for water applies:

Thus,

This shows a very small OH- ion concentration, but not zero, and illustrates the constant, reciprocal nature of H+ ions and OH- ions.

Example 1.24

A sample of lemon juice has a pH of 2.4. Find its [H+].

Solution

Recall that [H+] = 10-pH. Thus:

Example 1.25

Find, for a 0.014 M solution of slaked lime, Ca(OH)2:

A. [OH-]

B. pH

Solution

A. The subscript 2 next to (OH-) indicates that 1 mol of Ca(OH)2 produces 2 mol of OH- ions (along with 1 mol of Ca2+ ions).

It is assumed that Ca(OH)2 is a strong base.

Thus,

B. Furthermore, pOH = -log[OH-] = -log(0.028 M) = 1.55. Then,

Example 1.26: Acid-Base Reaction

A laboratory technician mixes 400 mL of a 0.125 M NaOH solution with 600 mL of a 0.100 M HCl solution. Find the pH of the resulting solution.

Solution

First compute the number of moles of acid and base to determine whether they are equal or, if not, which one is present in excess.

Since 0.06 mol of H+ >0.05 mol of OH-, the final solution will be acidic.

Furthermore, since 1 mol of acid reacts exactly with 1 mol of base, subtract moles of base from moles of acid to find the net moles of acid: 0.060 mol - 0.050 mol = 0.010 mol of H+ left over or unreacted and present after mixing. This amount is in a total, combined volume of 1,000 mL or 1.00 L. Thus,

Example 1.27: Acid-Base Titration/Neutralization

A laboratory technician wishes to find the concentration of an unknown base. He performs a titration in which 42.50 mL of 0.150 M HCl exactly neutralizes 25.00 mL of the base. Determine the concentration of the unknown base.

Solution

Since moles of acid (A) equal moles of base (B) at the endpoint (point of neutralization), use the relationship

Solve for MB:

Example 1.28: Dilution of a Solution

A laboratory technician is asked to prepare 500 mL of a 0.750-M solution of HCl. The stock solution of HCl that she has is labeled 6.0 M. How much (what volume) should she take from the stock solution bottle?

Solution

The total number of moles of HCl ultimately desired in solution is:

Moles of HCl = VhciMhci = (0.500 L)(0.750 mol/L) = 0.0375 mol

This is the amount that must come from the 6.0-M solution. Since this amount is simply being redistributed from a concentrated solution (Mi and Vi) to a more dilute one (M2 and V2), the number of moles of HCl in solution must remain constant. Thus, the number of moles of HCl before (1) must equal the number of moles of HCl after (2). So the calculation goes as follows:

Solve for Vi, substituting the values:

V = (0-750 M)(0-500 L) = 0.0625 L = 62.5 mL of stock solution 1 6.0 M

This same strategy can be applied to any dilution problem.

Supplemental Tables of Interest

Table 1.8. Common elements and their relative percent abundance in the Earthâ€™s (A) atmosphere, (B) hydrosphere, (C) biosphere, and (D) lithosphere

 (A) Relative Atomic Abundance of Elements in the Atmosphere Element Percent by Number (%) Nitrogen 78 Oxygen 21 Argon 0.93 Carbon dioxide 0.039 (B) Relative Atomic Abundance of Elements in the Hydrosphere Element Percent by Number (%) Hydrogen 66.2 Oxygen 33.2 Chlorine 0.3 Sodium 0.3 (C) Relative Atomic Abundance of Elements in the Biosphere Element Percent by Number (%) Hydrogen 49.7 Oxygen 24.9 Carbon 24.9 Nitrogen 0.3 (D) Relative Atomic Abundance of Elements in the Lithosphere Element Percent by Number (%) Oxygen 61.1 Silicon 20.4 Aluminum 6.3

(Continued)

 Hydrogen 2.9 Calcium 2.1 Sodium 2.1 Magnesium 2 Iron 1.5 Potassium 1.1 Titanium 0.2
 (E) Relative Atomic Abundance of Metals in the Lithosphere Metal Percent by Number (%) Aluminum 8.26 Iron 5.59 Calcium 4.12 Sodium 2.34 Magnesium 2.31 Potassium 2.07 Titanium 0.57 Other <0.50

## BIBLIOGRAPHY

Petrucci, R. H., Harwood, W. S., and Herring, F. G. 2002. General Chemistry, 8th ed. Upper Saddle River, NJ: Prentice Hall.

Chang, R. and Goldsby, K. A. 2013. Chemistry, 11th ed. New York: McGraw Hill.

Brown, T. L., Lemay, H. E., Bursten, B. E., Murphy, C. J., Woodward, P. M., et al. 2012. Chemistry, the Central Science. 12th ed. Upper Saddle River, NJ: Prentice Hall.

Lide, D. R., Ed. 2006. Handbook Chemistry and Physics, 87th ed. Boca Raton, FL: CRC.

Olia, M., Ed. 2013. Fundamentals of Engineering Exam, 3rd ed. Hauppauge, FL: Barron Educational Series.