Which statements are true about weak acids? They dissociate completely in water. The Ka values are typically more than 1000x smaller than 1. Weak acids correspond to a Kb expression. Usually the assumption method is okay to use if x/100<5% The assumption method allows (5.0-x) to become 5.0.
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- A solution containing 2.60 x 10-3 moles of NAOH and 3.32 x 10-3 moles of HCl into a container and then water is added until the final volume is 1.00 L. What is the pH of this solution? (Hint: pH + pOH=14; pH= -log1o [H"] and pOH = -log10 [OH"]) OA. 2.84 О в. 10.86 Oc 11.16 O D. 3.14 O E. 2.500.185 M in HCHO2 (Ka = 1.8 ×x 10-4) and 0.230 M in HC2H3O2 (Ka = 1.8 × 10-5) Express your answer to two decimal places. ? Find the pH of each mixture of acids. pH = 2.11 Submit Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining In a mixture of two acids that have similar K, values, the dissociation of one acid affects the dissociation of the other acid, and vice versa. Prepare two ICE tables (where I means initial, C means change, and E means equilibrium), one for the dissociation of HCHO2 and one for the dissociation of HC2H3O2. Use the given acid-dissociation constants to calculate the amount of H3O+ produced by each acid in the pair. Part D 6.00x10-2 M in acetic acid (Ka = 1.8 × 10-5) and 6.00×102 M in hydrocyanic acid (Ka = 4.9 × 10-10) Express your answer to two decimal places. ? pH =The acid dissociation constant K, of hypobromous acid (HBrO) is 2.3 x 10?. -6- a Calculate the pH of a 1.3 M solution of hypobromous acid. Round your answer to 1 decimal place.
- On acidity and basicity of aqueous solutions, the following relationships are valid, except If pH < 7, then [H+] < [-OH]; more acidic Ka x Kb = 1.0 x 10-14 [H+][-OH] = 1.0 x 10-14 =KW If pOH< 7, then [H+] < [-OH]; more basic pH + pOH = 14 for a weak acid and its conjugate base: pKa + pKb=pKW for a weak base: ↑Kb, ↓ pKb, more basic, more [-OH]14. Complete the following. a. Make a graph of the following data points. Is the relationship linear? If so, state the slope ume y-intercept. Are the variables directly proportional? Base Base pH pH (mL) (mL) 2.82 3.5 5.67 0.5 3.97 4.0 8.94 1.0 4.34 4.5 12.15 1.5 4.60 5.0 12.44 2.0 4.82 5.5 12.60 2.5 5.04 6.0 12.71 3.0 5.30 6.5 12.80 In some cases, the relationship between two variables is not linear, but the data can be manipulated to make a linear relationship. Using the data below, make a graph of the Volume vs. Pressure of a container, with volume being the independent variable. Is the relationship linear? If it is not, make a graph of Volume vs. 1/pressure. What would the pressure of the container be if the volume were 2.4 L? Show your calculation. Volume Pressure (L) (atm) 0.86 22.9 1.94 10.3 2.96 6.7 4.03 5.0 5.02 3.9 5.59 3.5The pH of a 0.0200 M solution of an unknown acid is 2.56. What is the Ka of this acid? To solve this problem: Write the acid dissociation equilibrium for the generic acid “HA” Set up an ICE chart ( with x = the concentration of H3O+ at equilibrium) Write the expression for Ka. Fill this in with the concentrations of H3O+, A- and HA at equilibrium, in terms of x. What is x? Can you find it from the given information? You should be able to use the pH to get the concentration of H3O+, which is x. Do this. Now that you know x, plug in into the Ka expression and find Ka.
- 7:30 1 y short X Powe X RI Anno x ! Ceng x PI TURUE A TINA A y weba x Piese y chem 101 Chem X i app.101edu.co Time's Up! Construct the expression for Ka for the weak acid, HCN. HCN(aq) + H.O(1) = H.O (aq) + CN (aq) Based on the definition of Ka, drag the tiles to construct the expression for the given acid. Ka = 5 RESET [H:O] [H.O'] [OH] [HCN] [H:CN') [CN] 2[H:O] 2(H.O) 2[OH] 2(HCN] 2[H.CN'] 2(CN] [H:OP [H.O'P [OHP [HCNJ" [H:CNP [CNP4. The equations and constants for the dissociation of three different acids are given below. HCO3 = H+ + CO?- K. = 4.2 x 10–7 -8 H;PO, = H+ + HPO- K. = 6.2 x 10 HSO, = H+ + SO?- Ka = 1.3 x 10–2 a. From the systems above, identify the conjugate pari that is best for preparing a buffer with a pH. of7.2. Explain your choice. b. Explain briefly how you would prepare the buffer solution described in (a) with the conjugate pair you have chosen. c. If the concentrations of both the acid and the conjugate base you have chosen were doubled, how would the pH be affected? Explain how the capacity of the buffer is affected by this change in concentrations of acid and base.H. | || The percent ionization of chloroacetic acid is less than that of fluoroacetic acid. :*-C-C-ö-H The percent ionizations cannot be compared without knowing the concentrations of the two acids. The structure of haloacetic acids, XCH,COOH (where X is The percent ionizations cannot be compared without knowing the pH of the solution. either F, Cl, Br, or 1), is shown above. The dissociation constants and molar masses of four haloacetic acids are listed in the table below. The percent ionization of chloroacetic acid is greater than that of fluoroacetic acid. Acid pK. K. Molar Mass(g/mol) CLEAR ALL Fluoroacetic acid 2.59 2.57 x 10–3 78.0 Chloroacetic acid 2.87 1.35 x 10–3 94.5 Bromoacetic acid 2.90 1.26 x 10-3 138.9 lodoacetic acid 3.18 6.61 × 10-4 185.9 An aqueous solution contains small but equal concentrations of both chloroacetic and fluoroacetic acids. Which statement comparing the percent ionizations of the two acids in the solution is true? エーO-
- *eL 24% Schoology 3:27 PM Tue Apr 12 < 88 Q A S22 1020 Unit 3 PS 1 - x Worksheet 15 Acid. X Worksheet 16 Wea. x Worksheet 19 Titr Planner Untitied (1) S22 1020 Unit.. 21020 Unit 1 P b. Calculate the pH of a 0.15 M pyridine solution. CSH&N(aq) CsH;NH" (aq) OH (aq) C[Review Topics] [References] Use the References to access important values if needed for this question. Consider the following equilibrium between 3-methylbenzoic acid (K,a-5.0x10) and its conjugate base: %3D CH3- CH3- H2O + H30* (1) Will lowering the pH shift the equilibrium to the right or to the left? (2) Calculate the ratio of the concentration of 3-methylbenzoate ion to the concentration of 3-methylbenzoic acid at a pH of 1.59. ratio [conjugate base] / [acid] Submit Answer Retry Entire Group 9 more group attempts remainingCalculate the pH of the following solutions. (1)Tap water after boiling (2) 0.0010 M NaOH aqueous solution (3) 10-9 M NaOH aqueous solution. (4) 0.20 M HAc (acetic acid, Ka(HAc) = 1.8 x10-5 ) (5) 0.20 M NaAc (sodium acetate). (6) 0.8203 g NaAc (FW=82.03 g/mol) was fully dissolved into 100.00 mL of 0.050 M HAc solution.