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pH calculations

pH calculations. strong acids/strong bases. pH scale. Below 7 is acidic Above 7 is basic or alkaline 7 is neutral. The controlling equilibrium in water solutions is: H 2 O + H 2 O  H 3 O + + OH - [H 3 O + ][OH - ] = 1 x10 -14

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pH calculations

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  1. pH calculations strong acids/strong bases

  2. pH scale • Below 7 is acidic • Above 7 is basic or alkaline • 7 is neutral

  3. The controlling equilibrium in water solutions is: H2O + H2O  H3O+ + OH- • [H3O+ ][OH-] = 1 x10-14 • If an acid is added to water, the [H3O+ ] increases while the [OH-] decreases. Their product stays the same (1 x10-14) . • If a base is added to water, the [OH-] increases while the [H3O+ ] decreases. Their product stays the same (1 x10-14).

  4. Find the pH of a .01M HCl solution Since the HCl is a strong acid, it dissociates 100% HCl + H2O  H3O+ + Cl- ; [H3O+ ]= .01M .01M = 1 x 10-2 pH = -log [H3O+] pH= - log (10-2) remember logarithms are exponents ….. So the pH = -(-2)= 2 Finding the pH if [H3O+ ] is exactly = 10 (whole number)

  5. Example 1 • Find the pH of a .001M solution of HNO3 • [H3O+] = .001 mol/L = 10-3M • pH= -log[H3O+] • pH= - log(10-3) • pH= -(-3) • pH= 3

  6. Finding the pH if [H3O+ ] is not exactly = 10 (whole number) • Find the pH of a .075M solution of HBr • HBr is a strong acid so the [H3O+] = .075M • pH = - log (.075)= -log (7.5 x 10-2) • pH= -(-1.12) = 1.12

  7. Example 2 • Find the pH of a .005M solution of HClO4. • HClO4 is a strong acid so [H3O+] = .005M • pH = - log (.005) • pH = -(-2.30) = 2.30

  8. Finding the pH if [OH- ] is exactly = 10 (whole number) • pOH is frequently used with bases. • pOH = - log [OH-] • pH + pOH = 14 for all water solutions • Find the pH of a .1M solution of NaOH • NaOH is a strong base so the molarity of the base=[OH-]. • pOH = - log (10-1) • pOH = -(-1) • pOH = 1 • pH = 14- pOH = 13

  9. Example 3 • Find the pH of a .0001M solution of KOH • Since KOH is a strong base, [OH-] = .0001M • pOH = - log (.0001) • pOH = -(-4) = 4.0 • pH=14- 4 = 10.0

  10. Finding the pH if [OH- ] is not exactly = 10 (whole number) • Find the pH of a .065M solution of LiOH • Since LiOH is a strong base, [OH-] = .065M • pOH = - log(.065) • pOH = -(-1.19) = 1.19 • pH = 14 – 1.19 = 12.81

  11. Example 4 • Find the pH of a .00325M solution of CsOH • Since CsOH is a strong base, [OH-] = .00325M • pOH = - log(.00325) • pOH = -(-2.49) = 2.49 • pH = 14-2.49 = 11.51

  12. Polyprotic acids or polyhydroxic bases. • Strong acids such as H2SO4 will dissolve in water and produce 2 moles of [H3O+] for one mole of acid. • So the [H3O+] of H2SO4 is twice the acid molarity • The pH of a .05M H2SO4 solution is …. • pH = - log (.1) • pH = -(-1) =1 • In a similar fashion, the [OH-] of a .02M Ba(OH)2 solution is twice the base molarity and is treated the same way as for a polyprotic acid. • pOH = -log(.04) = 1.39 • pH = 14-1.39 = 12.61

  13. pH to [H3O+] calculations • [H3O+] = inv log(-pH) = 10-pH • Find the [H3O+] of a solution whose pH=4.0 • [H3O+] = 10-4 M = .0001M • Find the [H3O+] of a solution whose pH=3.54 • [H3O+] = 10-3.54 = 2.88 x 10-4 M • Use the same procedure to find [OH-] from pOH

  14. Examples 5 and 6 • Example 5 • Find the [H3O+] of a solution whose pH=2.6 • [H3O+] = 10-2.6 = 2.51 x 10-3 M • Example 6 • Find the [OH-] of a solution whose pH= 5.8 • pOH=14-5.8 = 8.2 • [OH-] = 10-8.2 =6.3 x 10-9 M

  15. [OH-] or [H3O+] calculations • Ex. 1 – Find the [OH-] of a .25M HCl solution • [H3O+ ][OH-] = 1 x10-14 • .25 [OH-] = 1 x10-14 • 1 x10-14/.25 = 4 x 10-14 M

  16. pH calculations Weak Acids and Weak Bases • Weak acids such as acetic acid do not separate into ions 100% as do strong acids like HCl. • The pH of a 1.0 M HCl solution is 0.00 • The pH of a 1.0 M CH3COOH solution is 2.37

  17. Ka • The extent of ionization is measured by an equilibrium constant (Ka) • If you use HA as the formula for a generic weak acid HA + H2O   H3O+ + A- • The Ka= [H3O+ ][A-]/[HA] • The [ ] represents molarity • For CH3COOH (acetic acid) Ka = 1.8 x 10-5

  18. R I C E table and Ka • The method used to calculate pH from Ka is called rice for • Reaction • Initial conditions • Change • equilibrium

  19. Find the pH of a .50M CH3COOH solution • R CH3COOH + H2O   H3O+ + CH3COO- • I .50 M 0.0M 0.0M • C -x +x +x • E .50-x x x • Then set up the Ka equation • 1.8 x 10-5 = [H3O+] [CH3COO-]/[CH3COOH] • 1.8 x 10-5 = x2/(.50-x)

  20. Approximation method • We will always decide that the –x in the denominator of 1.8 x 10-5 = x2/(.50-x) is too small to make a significant difference in the calculation so the equation simplifies from a quadratic equation to • 1.8 x 10-5 = x2/.5 • X= (9 x 10-6)1/2 • x = .003M • pH = -log (.003) = 2.52

  21. Weak base calculations use Kb • The method of solving a Kb question is the same as a Ka question • Example: Find the pH of a .450M NH3 solution if the Kb = 1.8 x 10-5 • R NH3 + H2O  NH4+ + OH- • I .450M 0 0 • C -x x x • E .45-x x x • 1.8 x 10-5 = [NH4+] [OH-] / [NH3] • 1.8 x 10-5 = x2/(.45-x)   1.8 x 10-5 = x2/.45 • X = 2.85 x 10-3 M • pOH = -log(2.85 x 10-3 ) = 2.55 • pH = 14 –pOH = 14- 2.55 = 11.45

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