Chapter 15 Acids and Bases Notes PDF

## Chem 102: Chapter 15: Acids and Bases ### Bronsted Acid Proton donor ($H^+$) ### Bronsted Base Proton acceptor ($H^+$) ### Conjugate Base of the Bronsted Acid What remains when $H^+$ is donated- basic ### Conjugate Acid of the Bronsted Base what forms when $H^+$ is accepted. | Bronsted Acid...

## Chem 102: Chapter 15: Acids and Bases ### Bronsted Acid Proton donor ($H^+$) ### Bronsted Base Proton acceptor ($H^+$) ### Conjugate Base of the Bronsted Acid What remains when $H^+$ is donated- basic ### Conjugate Acid of the Bronsted Base what forms when $H^+$ is accepted. | Bronsted Acid | C.A. | Conjugate Base | B.B | | :-------------: | :---: | :--------------: | :---: | | HI | | I- | | | $H_2SO_4$ | | $HSO_4^-$ | | | $H_3PO4$ | | $H_2PO_4^-$ | | *diprotic acid - 2 H last in 2 steps* *(triprotic)* | Bronsted Base | C.B. | Conjugate Acid | B.B | | :-----------: | :---: | :------------: | :-: | | $NH_3$ | | $NH_4^+$ | | | $OH^-$ | | $H_2O$ | | *ammonia* ### Bronsted Acid-Base Reaction: $NH_3 + H_2O \leftrightharpoons NH_4^+ + OH^-$ BB B.A C.A C.B $H_3O^+$: hydronium ion "hydrated $H^+$" ### Water "the universal solvent" *(can dissolve alot)* - It can act as an acid or a base - Undergoes auto ionization $H_2O_{(l)} + H_2O_{(l)} \leftrightharpoons H_3O^+_{(aq)} + OH^-_{(aq)}$ BB B.A C.A C.B ### Ion Product Constant for Water $K_w = K_c = [H^+][OH^-] = 1 \times 10^{-14}$ *Ion product constant for water at 25°C* The $[H^+]$ of an HCl solution is 1.3 M. Calculate the $[OH^-]$. $[H^+] \times [OH^-] = 1.0 \times 10^{-14}$ $(1.3)\times[OH^-] = 1.0 \times 10^{-14}$ $[OH^-] = 7.7 \times 10^{-15}M$ *Is this solution acidic, basic, or neutral?* $[H^+] > [OH^-]$ acidic Soren Sorensen invented pH in 1909 $pH = -log[H^+]$ pH: "power of $H^+$ $pH + pOH = 14.0$ If the $[H^+]$ of a sample is $7.5 \times 10^{-4} M$, calculate the pH and the pOH. $[7.5 \times 10^{-4}][OH^-] = 1 \times 10^{-14}$ $[OH^-] = 1.33 \times 10^{-11} M$ $pOH = -log [OH^-]$ $pOH = -log(1.33 \times 10^{-11}) = 10.88$ $14-10.88= pH = 3.12$ -> pH is in range cus its acidic The pH of a sample of acid rain was 5.21. Calculate $[H^+]$and $[OH^-]$ $pH = 5.21 = -log[H^+]$ $pH = -5.21 = log[H^+]$ $10^{-5.21} \times 6.17 \times 10^{-6} M = [H^+]$ $pOH = 14 - 5.21 = 8.79 = log [OH^-]$ $1.62 \times 10^{-9} M = [OH^-]$ ## Chapter 15: Acids and Bases Part II * Write out what names * Dissociate * Calc ### Strength of acids and bases: *Strong" completely dissociates into ions in solution. ### Strong acids: $HCl, HBr$ $H_2SO_4, HClO_4$ $HI, HNO_3$ #### Example $HBr + H_2O \to H_3^++ Br^-$ $HBr+H_2S \to H^++ Br^-$ $HBr_{(aq)}\to H^+_{(aq)} + H^-_{(aq)}$ ### Weak acids: partially dissociates into ions in solution. #### Example $HF_{(aq)} \leftrightharpoons H^+_{(aq)}+F^-_{(aq)}$ $HNO_2$ ### Strong bases: Group I hydroxides $Ba(OH)_2$ *Strong Base* $RbOH_{(aq)} \to Rb^+_{(aq)} + OH^-_{(aq)}$ ### Weak bases: $CuOH_{(aq)} \to Cu^+_{(aq)} + OH^-_{(aq)}$ Calculate the pH of a $4.0*10^{-3}M$ $HNO_3$ solution. $4 \times 10^{-3}M$ $HNO_3$ $HNO_3 \to H^+ + NO_3^-$ *Reactants:* Strong- subtract all Weak- we dunno Initial 4.0x10-3 $\Delta = -4.0 \times 10^{-3}$ final 4.0 x 10-3 *Calc H+* $pH = -log (4.0 \times 10^{-3}) = 2.40$ ### Weak acids and ionization constants/dissociation constants Given a weak acid, HA: $HA_{(aq)} \leftrightharpoons H^+_{(aq)} + A^-_{(aq)}$ $K_a = \dfrac{[H^+][A^-]}{[HA]}$ - acid dissociation constant Large $K_a$ = more dissociation occurs Given $k_a$ means weak What is the pH of a 0.50 M HF solution at 25°C? $K_a$ for HF = $6.3 \times 10^{-4}$ $HF \leftrightharpoons H^+ + F^-$ Initial 0.5 0 0 $\Delta$ - x + x+ x -> 0.50-x x x $K_a = 6.3 \times 10^{-4} = \dfrac{x^2}{0.50-x}$ $x^2 + 6.3 \times 10^{-4} x - 3.15 \times 10^{-4}$ ### Method of successive approximation: Assume x is $<<$ smaller than 0.50 $0.5 \approx 0.50$ $6.3 \times 10^{-4} = \dfrac{x^2}{0.50}$ $x = 1.77\times 10^{-2}$ ### Check if really smaller than 0.50 $\dfrac{1.77 \times 10^{-2}}{0.50} \times 100 = 3.54$% within 5% assumption valid A) Calculate the pH of a 0.01 M $HNO_2$ solution B) Calc pH of a 0.015 M $HNO_3$ solution. *Example* ## Calculate the pH of a 0.72 M $NH_3$ solution. $K_b = 1.8 \times 10^{-5}$ $NH_3 + H_2O \leftrightharpoons NH_4^+ + OH^-$ Initial | 0.72 | 0 | 0 :----: | :----: | :----: | :----: $\Delta$ | --x | x | x |0.72-x | x | x $K_p$ = base. Assume $x<<0.72$ $x = 0.0036$ $1.8 \times 10^{-4}$ *Check* pH = 2.44 $pH = 14 -2.44 = 11.56$ What is the ptt of a 0.50M HF solution at 25°C Method of Successive approximation *Substitute x into $k_a$:* $6.3 \times 10^{-4} = \dfrac{x_2}{0.05-5.3 \times 10^{-3}}$ Solve for x: $\times 10^{-4} = \dfrac{x_2}{0.0444}$ Solve for x: $(x = 5.3 \times 10^{-3} M)$ Not the same *Repeat Substitute* $x$ into $k_a$ $6.3 \times 10^{-4} = \dfrac{x_2}{(0.05-5.3 \times 10^{-3})}$ Solve for x: $6.3 \times 10^{-4} = \dfrac{x_2}{0.0447} = 5.3 \to \times 10^{-3} M$ *So, we assume* that is $x$. 2 answers are same to 2 sig figs! -*next step* -*log(5.3 x 10-3) = pH = 2.28* $HNO_3 \leftrightharpoons H^+ + NO_3$ Initial 0.015 0 0 $\Delta$ 0.015-X X X final. 0.015-x x x $H^+$ = -log(0.015) = 1.8239 2 Ka = 5,6 x 10-4 =x² 0,015 -X 5.6 x 10-4 (0.015-x) = x² 2 -x² 7.74 x10-6 -5.6x10 x = 0 2 -x²- (5.6x\10-4x) + (7.74×10-6)=0 - b±√b2-4ac -------------- - 2a 5.6 x 104 + √(-5.6×10-4) 2-4(-1) (7.74x10-6) ------------- -2 -2.8×10 ± 0.055922804 ------------- -2 -2.8 x 10"± -0.0027961402 $\times= - \dfrac{0.0030761402}{2.52 x 10^{-3}} $ = -20% $\times= - \dfrac{0.0025161402}{4} = 16$% -> \$- $\log(2.52×103)$ 2.59 = pH ### The pH of a 0.20 M acetylsalicylic acid solution is 2.12. What is its $K_a$? *aspirin* $CaH_8O_4 \leftrightharpoons H^+ + CaH_7O_4$ Initial 0.2 0 $\Delta$ -x + x+ x 0. 2-x O + X O + X *(Relationship between Ka and Kh)* High Ka % ionization high $\dfrac{X^2}{O.2-X}-what$ is \$-log[H*] = 2.12 10g[H+] = -2.12 10^ [H] = $1.9 \times 10^{-3}$ *Sce next pare* $K = \dfrac{\\(7.59 \times 10^{-3})^2\\}{(0.2 - 7.59 \times 10^{-3}}$ $K = 3.0 \times 10^{-4}$ ### %Ionization % ionization= ionized acid concentraion at equilibrium/ Initial concentration of acid*100 *Diprotic/Triprotic acids:* *(acids ability to lose 𝐻^+ depend on structure)* test - only monoprotic # Calculate the final concentrations of a0.10 M $C_2H_2O_4$ solution. *Diprotic acid:* $C_2H_2O_4$ *P1:C_2H_2O_4 \leftrightharpoons H++C_2HO_4$ initial | 0.10 | 0 | 0 :----: | :----: | :----: | :----: $\Delta$ | --x | +x | + x | 0.10- x | x | x *Ka_1 = 6.5 × 10^{-2}=\dfrac{ X^2}{0.10-X}$ *Solve \to* * [H+] = [C₂HO₄] = 0.054 M* [$C_2H_2O_4$] = 0.10 - 0.054 # Step 2 $C_2HO^+ \leftrightharpoons H^+ + C_2O_4^{-2}$ initial | 0.054 | .054 | 0 :----: | :----: | :----: | :----: $\Delta$ | -y | +y | +x 0. 54-y/.054+7 Shortcut assume $y<<0.054$ $.054-Y \approx 0,054$ $Y = 6. \times 10^{-5}$ $5-LESS E makes strongest$ $More BDE Weakest$ diagrees w/ polarity Strength of Acids Based on Molecular Structure *Strong bond* = Less dissociation = weaker acid. Bigger difference in electronegative 1 more $H^+$ ionizing, $H^+$easier to Break! 2 factors that influence the amount of ionization: 1 Strength of H-X bond 2. Polarity of H-X bond Elestro Hydrohalic acids- acids that contain H and halogen neg 2. |5 $HF << Hbr < HCL < HI$ $HF 568.2$ - Weak 298.3 strong Oxoacids acids that contain H.O. and another element typei 2 types 1. Different central atom from the same group $(HCIO_3)$ vs $(HBrO_3)$ 2. Same central atom, different number of attached groups $HCIO_4 HCIO_3 HCIO_2 $ Again look @ Polarity O is an electron withdrawing group. Acid-Base Properties of Salts: *ACID + BASE-> Salt + water* all Salts are ionic compounds *Salt hybrolysis* the reaction of an onion or cation of salt (or both) with water to figure out what type of sall? Cation from a strong base will NOT react with water ${Na ^+\; Na} others$ 1 Strong acid stroug base $\approx HNO_+} KOH \approx (NaBr)_k < Ho$ Hydrolysi reaghin tells us its an andic salt? releasig H weak base 4. weak Acid +weak Base $\approx HF+ NHyDH \approx (NHyF)_s = Hoo$