Stuff Make-up Exam tomorrow night, Date: Tuesday Feb 17. Room: C-109 Time: 6-8 (1.5 hr) Problems for Acid-Bases posted. Hour Exam Solutions posted Lecture slides up to date Can I ask the girls to email what high school you attended. Thank you!
Kw is an equilibrium constant that describes the autoionization of water. H 2 O(l) H + (aq) + OH (aq) K w = [H + ][OH - ] H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH (aq) Base Acid Acid Base K w = [H 3 O + ][OH - ] When we measure equilibrium concentrations of OH and H + we find that: K w = [H3O + ][OH - ] = 1 X 10-14 Called the Ion-Product of Water
ph is the human defined logarithmic scale for measuring acidity. It is directly related to [H + ] concentration in an aqueous solution. ph = -log [H + ] = -log [H3O + ] logarithmic form p = -log [H + ] = 10 -ph exponential form ---Each ph unit or number represents a power of 10 (or an order of magnitude as it is called). A solution with a ph of 4 is 10X more than one with ph=3, 100X more than ph=2 and 1000X more than ph = 1.
Acidic, basic and neutral solutions are identified by their ph values [H3O + ] ph [OH - ] ph = -log [H3O + ] K w = [H3O + ][OH - ] = 10-14 The ion-product of water is a constant!
ph is measured with dyed paper or with a commercial instrument. ph meter ph (indicator) paper
Known dye indicators change color within narrow ph limits.
Kw is related to ph via logarithm. K W = [H 3 O + ][OH - ]= 1.0 X 10-14 take -log of both sides -log Kw = -log ([H 3 O + ][OH - ]) = -log(1.0 X 10-14 ) -log K W = - log[h 3 O + ] + -log[oh - ] = -log (10-14 ) pk W = ph + poh = -(-14) pk W = ph + poh = 14 K W = [H 3 O + ][OH - ] = 10-14 Connect The Dots: Kw---pH---pOH
[H3O + ] ph [OH - ] poh Relation ships between [H 3 O + ], ph, [OH - ] and poh
A Logarithm is Just An Exponent Let y be any number. Also, let b be a number and let s call it the base, b. Lastly let s call the power we raise the base to an exponent and label it as x y = b x 100 = 10 x b = base, x = exponent and y = a number b = 10 and y =100 and x = exponent The logarithm of 100 {log (100)} is the value of the exponent, x, needed to raise 10 to in order to obtain 100. log (100) = x = 2 The logarithm of 100 is value of the exponent required in the equation above!
Logs are easy if you see the riff. What is the log10 of the following numbers? 1000, 100, 10, 1, 0, 0.1, 0.01, 0.001 log10 1000 = log 1000 = x log10 100 = log 100 = x log 10 = x log 1 = x log 0 = x log 0.1 = x log 0.01 = x Determining the log 10 is the same as computing x in this equation! means 10 X = 1000 10 X = 100 10 X = 10 10 X = 1 10 X = 0 10 X = 0.1 10 X = 0.01
In math we use two common logarithm functions. Log to the base 10 and the natural log. log = log 10 log to the base 10 log e = log 2.7183 = ln the natural log Properties of Logs log (A B) = log A + log B ( ) A log = log A log B B log A n = n log A antilog(log x) = 10 log x = x
Antilog of a number means use the number as an exponent. Sometimes we know the base and the exponent, but we need to compute the number. We can write it like this: log (?) = 2 In words we would say: The log10 of what number is equal to 2? Finding this number is called taking the antilogarithm Antilog [log (?)] = Antilog (2)? = 10 2 = 100 The antilog 10 is the same as the base 10 operator!
Log and Exponential Forms Logarithmic Form ph = -log [H 3 O + ] Exponential Form 10 -ph = [H 3 O + ] poh = -log [OH - ] 10 -poh = [OH - ] pk a = -log K a 10 -pka = pk a When H 3 O +, OH -, K a are large values----the pk s are small values. Strong acids have large H 3 O and K a therefore small ph and small pk a
The ph of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H + ion concentration of the rainwater? The OH - ion concentration of a blood sample is 2.5 x 10-7 M. What is the ph of the blood?
The ph of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H + ion concentration of the rainwater? ph = -log [H + ] antilog(-ph) = antilog (log [H + ]) [H + ] = 10 -ph = 10-4.82 = 1.5 x 10-5 M The OH - ion concentration of a blood sample is 2.5 x 10-7 M. What is the ph of the blood? ph + poh = 14.00 poh = -log [OH - ] = -log (2.5 x 10-7 ) = 6.60 ph = 14.00 poh = 14.00 6.60 = 7.40
Problem: Calculating [H 3 O + ], ph, [OH - ], and poh In a restoration project, a conservator prepares copperplate etching solutions by diluting concentrated nitric acid, HNO 3, to 2.0 M, 0.30 M, and 0.0063 M HNO 3. Calculate [H 3 O + ], ph, [OH - ], and poh of the three solutions at 25 o C. PLAN: HNO 3 is a strong acid so [H 3 O + ] = [HNO 3 ]. Use K w to find the [OH - ] and then convert to ph and poh.
In a restoration project, a conservator prepares copperplate etching solutions by diluting concentrated nitric acid, HNO 3, to 2.0 M, 0.30 M, and 0.0063 M HNO 3. Calculate [H 3 O + ], ph, [OH - ], and poh of the three solutions at 25 o C. For 2.0 M HNO 3, [H 3 O + ] = 2.0 M, and ph = -log [H 3 O + ] = -0.30 = ph [OH - ] = K w / [H 3 O + ] = 1.0 x 10-14 /2.0 = 5.0 x 10-15 M; poh = 14.30 For 0.3 M HNO 3, [H 3 O + ] = 0.30 M and -log [H 3 O + ] = 0.52 = ph [OH - ] = K w /[H 3 O + ] = 1.0 x 10-14 /0.30 = 3.3 x 10-14 M; poh = 13.48 For 0.0063 M HNO 3, [H 3 O + ] = 0.0063 M and -log [H 3 O + ] = 2.20 = ph [OH - ] = K w / [H 3 O + ] = 1.0 x 10-14 /6.3 x 10-3 = 1.6 x 10-12 M; poh = 11.80
Equilibrium Problems Strong Acids/Bases Weak Acid/Bases 100% dissociation of [HA] % ionization varies with concentration of acid. [H 3 O + ] = [HA]. No need for ICE table. [H 3 O + ] = [HA]. ICE Table Is Needed 1. Simplification If 100 x Ka < [A]0 then [HA] - x = [HA] 2. Perfect square 3. Quadratic equation
Know your strong acids and bases and you will recognize the problem hydrohalic acids perchloric nitric sulfuric Group I hydroxides Group II hydroxides
Acid-Bases Equilibrium Problems HCl(aq) + H 2 O H 3 O + (aq) + Cl (aq) 0.004% at equilibrium 99.996% at equilibrium STRONG ACID CH 3 CO 2 H(aq) + H 2 O H 3 O + (aq) + CH 3 CO 2 (aq) 98.7% at equilibrium 1.3% at equilibrium WEAK ACID