DAT acid and base general chemistry

Consider the following 1 Molar aqueous solution of acetic acid (chemical formula here) K a = 1 x 10 to the -5. What would be the pH of the solution at equilibrium?

shortcut: ph = negative log square root of (Molarity) (ka)

Which amount and concentration of base would neutralize 100 mL of 0.1 M sulfuric acid?

-100ml of 0.2M NaOH -Sulfuric acid →"ic" one of the strong acids--> H2 SO4→ 2 ions -(2)(0.1 mol/L)(0.1 L) = nA MAVA -Use MA to find MB → 0.02 mol = MB -Diprotic acids, such as sulfuric acid (H2SO4), carbonic acid (H2CO3), hydrogen sulfide (H2S), chromic acid (H2CrO4), and oxalic acid (H2C2O4) have two acidic hydrogen atoms. -Triprotic acids, such as phosphoric acid (H3PO4) and citric acid (C6H8O7), have three.

A. Br- B. Na3PO4 C. NH3 D. Cl- E. H2O Match 3 M with 3negative charges We need 3 neg charges, if we dissociate the molecules given Na3PO4 is able to absorb 3 H+ protons, because it dissociates into the very basic PO43- ion.

A researcher has a solution of 3M nitric acid that she needs to neutralize. In order to do this, she decides to add the acid to a solution to neutralize it. Which of the following would neutralize the acid the fastest?

A. 10-pOH B. 10-(14-pOH) C. 10(14-pH) D. 10-(14-pH) E. 10(14-pOH) [H+] = 10 (-pH) pH + poH (4.12) =14

An unknown solution is found to have a pOH of 4.12, which of the following would be equal to the [H+] of the solution? t3

C. 11 --find mols of OH using given molarity -we are only given Barium OH info -GMR any time given the rxn→ use Mols and Ratio from balanced equation to find [OH] mols -Use BaOH [.0005Mol/L] → 2 OH/1 BaOH molar ratio→ to find OH [Mol/L] (use dimensional analysis) .-001 M → [1 x 10 -3 ] = [OH] -OH → [-log OH]--> - log (1 x 10 -3) ⇒ 3 -pOH + pH = 14 → 14-3= 11pH

Barium hydroxide dissociates into ions, as shown in the reaction below. If there is 0.0005M barium hydroxide, what is the pH of the solution? t6 1 Ba(OH)2(aq) →Ba2+(aq) + 2OH-(aq) A. 12 B. 10 C. 11 D. 2 E. 3

The following mixtures (assume volumes are additive) will result to a buffer solution EXCEPT one. Which one is the EXCEPTION?t6 A. 100 mL 0.10 M CH3COOH + 100 mL 0.10 NaCH3COO B. 100 mL 0.40 M HCN + 100 mL 0.20 M KOH C. 100 mL 0.050 M HCl + 100 mL 0.075 M NaOH D. 100 mL 0.125 M NH4Cl + 100 mL 0.100 M NH3 E. 100 mL 0.100 M HCl + 100 mL 0.20 M NH3

C. 100 mL 0.050 M HCl + 100 mL 0.075 M NaOH "Except" = means we want 3rd rule as answer= look for strong acid and strong base Ways to get a buffer" 1:1 ratio (weak/strong) 2:1 (weak/strong) Never (strong/ strong)

-2.0 x 10 -1 Given M and mL→ n1M1V1= n2M2V2 Equivalence point → titration→ n1M1V1= n2M2V2 M1(20 mL) = 2(0.05 M)(40 mL) M1 = 2(0.05 M)(40 mL) / 20 mL = 0.2 M HCl or( 2×10-1 M HCl)

Calculate the molarity of a 20.0 mL sample of HCl given that 40 mL of 0.05 M Ba(OH)2 was added to reach the equivalence point.

Which statement explains why HI is a stronger acid than HBr, which is a stronger acid than HCl?

D. The bond strength between H-I is lower than that of H-Cl or H-Br. -the less electronegative the more available the H bond

Consider the following 1.3 M aqueous solution of acetic acid: HC2H3O2 (aq) -> H+ (aq) + C2H3O2 (aq), Ka = 1.8 x 10-5 What would be the pH of the solution at equilibrium? Test

Given M and Ka→ weak acid formula = [H] [H^+] = √(1.3)(1.8 x 10^-5) Next, to determine the pH, the -log must be taken of the [H+], giving the final answer of:-log √ (1.3) (1.8 x 10^-5) -log ([x]2 = 1.3*1.8x10-5) E.

hydrobromic acid formula

HBr one of the 7strong acids

Ka x kb = 1x10 -14 we would write: -14 - 9 = -23. // A. 1x10-23

If Ka = 1x10^9, what is Kb? A. 1x10-23 B. 1x10-5 C. 1x105 D. 1x1014 E. 1x1023

Since pKa + pKb = 14, and pKb is given as 1.4, then pKa must be 12.6. Higher pKa's correspond to weaker acids, so B is the correct answer. B. Weakly acidic Conj base → pkb 1.4→ "weak" base→ FLIP Molecule in question → "WEAK" ACID -Think "pkb" = switch the scale,

If the conjugate base of a molecule has a pKb of 1.4, what would you expect the molecule to be? Strongly acidic B. Weakly acidic C. Neutral D. Weakly basic E. Strongly basic

Answer:E Write it out and name each one Base is the one that accepts the H

Look for the one that carries over the H+ bigger number exponent will be base

Given that the Kb of CH3NH2 is 1.8 x 10-6, what is the pH of a 0.2 M aqueous solution of methylamine? C. 7.3 E. 12.3 D. 10.8

--Kb and M → weak acid formula = [OH] -all decimals to exponents→ 1.8= 18 / 0.2 = 2 -Sqrt[ (18x10 -7) (2x10 -1)]--> (6x10-4)= [OH] -Log (6x10-4) = poH = 5 (above halfway point) We want pH → 14 - pOH = pH 14-5 = 9 C. 7.3 E. 12.3 Closest to 9 is 10.8 D. -Solving logs: -Log (1x10-6) = 6 -Log (3.6 x10-6) = 6.5 (halfway point)

Ways to get a buffer"

-1:1 ratio (weak/strong) -2:1 (weak/strong) -Never (strong/ strong)

Adding sodium oxalate to a solution of oxalic acid causes the pH to:

A. increase due to the common ion effect. Whenever you're using a weak acid, like oxalic acid, write the equation as: HA <--> H+ + A- HA is oxalic acid, the H+ is what makes something acidic or basic. The A- is the oxalate. By adding oxalate, a common ion, you push the equation to the left. In the process, you decrease the amount of H+, making the solution more basic.

Ka x kb formula

Ka x kb = 1x10 -14

look for the concentration [M] to get pH "What is the pH" → given naOH VM → pH+poH =14 "Adding" → V2 = (V1 + V2) V1 M1 = (V1+V2) (M2) (0.02 L)(0.02 M naoh) =(0.04 L)(M) → 0.01 M2 =[OH new] .0004/.04 →( 4.x10-4 / 4x10-2 )→ -4 - -2 = -2 = M2 Convert M2 to sci not. → .01 → 1x10-2 = [OH] Also can do it this way: pOH = - log (1x10-2) = poH = 2 14 = pOh + pH (has the p's in this formula) 14- 2= 12 pH

What is the pH of a solution prepared by adding 20 mL of 0.02 M NaOH(aq) to 20 mL of water? A. 2 B. 3 C. 10 D. 11 E. 12

What is the concentration of [CN-] in a 0.2 M solution of HCN? The Ka of HCN is 5 × 10-10. A. 1 × 10-5 B. 1 × 10-10 C. 1 × 10-11 D. 5 × 10-5 E. 5 × 10-10

move the decimal points to get whole numbers after multiplying move them back [CN-] = √10 x 10^-11 [CN-] = √1 x10^-10 when you square root an exponent you divide your exponent by 2 [CN-] = 1 x 10^-5 A. 1 × 10-5

If the Ka of acetic acid is 1.8 × 10-5, what is the pKb of acetate?

pKa + pKb = 14 pKb = 14 - pKa pKb = 14 - (-log(Ka))pKb = 14 - (-log(1.8 × 10-5)) The two negatives cancel to form our final equation.pKb = 14 + log (1.8 × 10-5)

Which of the following concentrations of hydrobromic acid would yield a solution with PH 3.0? 1x10-3 mM 3x10-2 mM 1x10^0 mM

ph --> OH conversion OH = 10^ -ph the answer is redundant; dont let it throw you off 1x10^0 mM is the same as 1 mM 1mM --> .001 M --> 1x10^-3 -1x10^0 mM is the answer

What ratio of sodium acetate concentration to acetic acid concentratoin is needed to make a buffer of ph 4.8 (Nore the pka of acetic acid is 4.8.

ph = pka if ratio of acid and base are equal

Which of the following is the strongest acid?

the more O's the more pull leaving the H vulnerable to be used

Which of the following molecules has the highest pKa?

-CH4 High pka = weak acid inverse relatinship "p" alwats reverses things ka would be directly correlated to the acids strength ONLY when "conjugate base" is in the question does the scale switch

What is the effect when NaF is added to an aqueous solution of HF? Boost 3 A pka BThe pH of the solution would increase C D E ka

-NaF will dissociate into F- → shifting to make more HF. -This means the H+ and F- in solution will come back together, forming more HF. Therefore, the concentration of H+ in the solution will decrease, resulting in a solution with a higher pH. -Ka and pKa are constants. These two constants only change in relation to temperature. We can rule out answer choices [A] and [E].

pyruvic acid is a weak acid. Which of the following is a likely pH for an aqueuos solution of sodium pyruvate.

-pyruv-ate is the key word--> their focusing on the base part of the solution (don't get thrown off by the first sentence) -its the base part meaning the solution will be basic -pyruvate is not a known strong base so the pH will not be close to 14 it will be closer to 8 the answer is not 6 because that's below 7 in the acid range and were NOT focusing on acid answer is not 7 bc that's neutral neither acid/base

Calculate the molarity of a 2L sample of HCl given that 4L of 5 M Barium Hydroxide was added to reach the equivalence point.

1. Equivalence point is key word 2. use n1 MV = n2 MV 3.plug and chug

When titrating the monoprotic acid HCl with NaOH of a known concentration, which of the following must be equal at the "equivalence point"?

1. Initial moles of HCl are equal to the moles of OH- added 2. Only moles affect equivalence point

E. Water acts as an acid in reaction 2, and a base in reaction 1 Find the most acidic molecule to the LEFT in each reaction (1) NH4 is strongest, leaving water as base (2) NaH vs. H2o → neither is a strong acid, dissociate into ions and compare O2- vs Na+ = O is more electronegative than Making the H+ in H20 more acidic

Two acid/base reactions are shown below: Which of the following is true?

answer b

Which of the following compounds should have the strongest conjugate acid? (See choices in answer).

-D. HClO4 -go with what you know the more O's the more pull leaving the H vulnerable to be used by bases= most acidic

Which of the following is the strongest acid? A. HClO B. HClO2 C. HClO3 D. HClO4 E. ClO4-

Answer is D: acidity and size/ electronegativity -the less electroneg. The more available the H can bond -Iodine molecule is bigger making H-I interaction weaker and making H+ more available as an acid

Which statement explains why HI is a stronger acid than HBr, which is a stronger acid than HCl? t2 A. Iodine is the smallest element out of the 3 and thus creates the strongest acid. B. The acid strength is increased as electronegativity increases in the binding element. C. The number of available electrons in iodine for binding is greater than that of Cl or Br. D. The bond strength between H-I is lower than that of H-Cl or H-Br. E. Acid dissociation decreases as the shielding effect increases.