1. Prediction of pH and hydrolysis of salts: Aqueous Solutions NH4CI NaCI Na2CO3 CUNO3 NaC2H3O2 NaHSO3 Nat Nat Nat Nat Cation cut co3 2- Anion pH Prediction =7 <7 (<7, >7 or =7) Cation and Anion Hydrolysis Reactions: lon Acidic/Basic/Neutral Hydrolysis Reaction NH4* acidic neu tral CI- Na+ neutral CO32- basic + H40ce) HCO3 caa) + OH"cae) Cut basic Co* + thole) > cU OH +H* NO3 neutral C2H3O2" kg- 1.0xlo HSO3 HSO3 +H20 14) 803caq,) -14 acidic 10 1-SxI07 HISO3 caq) + H20ce) 2 1250s + 0 t caas Kb= ka - 6.7 da kazke

1. Prediction of pH and hydrolysis of salts: Aqueous Solutions NH4CI NaCI Na2CO3 CUNO3 NaC2H3O2 NaHSO3 Nat Nat Nat Nat Cation cut co3 2- Anion pH Prediction =7 <7 (<7, >7 or =7) Cation and Anion Hydrolysis Reactions: lon Acidic/Basic/Neutral Hydrolysis Reaction NH4* acidic neu tral CI- Na+ neutral CO32- basic + H40ce) HCO3 caa) + OH"cae) Cut basic Co* + thole) > cU OH +H* NO3 neutral C2H3O2" kg- 1.0xlo HSO3 HSO3 +H20 14) 803caq,) -14 acidic 10 1-SxI07 HISO3 caq) + H20ce) 2 1250s + 0 t caas Kb= ka - 6.7 da kazke

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter16: Applications Of Neutralization Titrations

Section: Chapter Questions

Problem 16.34QAP

See similar textbooks

Similar questions

Ka for ethanoic acid is 1.7*10^-5 moldm-3. Show by calculation that the initial pH in experiment B is 4.8.
Acetic acid (CH3COOH, abbrev. HA) is a very common weak acid with pKA = 4.75 (at 25oC and low ionic strength). Calculate pH of the diluted solution of acetic acid with concentration c = 0.05 mol dm-3.
ALL INFORMATION NEEDED IS PROVIDED, I NEED FIRST QUESTIONS ANSWERED FIRST BEFORE OTHERS CAN BE ANSWEREDRate = -(Delta[RCl])/Delta t andln[RCl]t=-k*t+ln[RCl]0 < [R-Cl]-time equation Test tube A = 0.30 of 0.10M NaOH(aq), 6.70ml DI H2O & 1 drop of bromophenol bue Test Tube B = 3.00M of 0.10M R-CL in acetone. Mix together in 3 water baths Bath A Trial 1= 20C, 53.6s Trial 2= 20C, 49s for color to changeBath B Trial 1= 32C, 18s Trail 2= 32C, 20s Bath C Trial 1= 10C, 204s Trial B= 10C, 200s average out both trials for each bathcolor change happens after 10% of R-Cl has reacted so In terms of R-Cl this means that 10% of the R-Cl has reacted and ____ of the R-Cl is left Q1. at the time of the color change [R-Cl]t=____[R-Cl]o (Fill in the blank) Q4. Substitute the value of [R-Cl]t into appropriate equation listed above and solve for k in terms of [R-Cl] and t. Q5. Use average time for each temp and determine the value of k (with units) at each temperature Q6. Rearrange the Arrhenius…
At 25oC, Ka for acid dissociation of aspiring (C9H8O4) is 3.0 x 10-4. Calculate ΔGo for the reaction C9H8O4 (aq) + H2O(l) H3O+(aq) + C8H7O4-(aq)
(9.29 × 105/8)-20.81 = with correct sig figs
1. is pH scales is strictly btw 0 and 14? if not, why? 2. what is CSTR? is this heterogenous? is this concentration depends on location? 3. is steady state, mass can enter ot not? leave the control volume or not?
MATLAB The acidity of solutions is generally measured in terms of pH. The pH of a solution is defined as theconcentration of hydronium ions. Create a table of conversions from concentration of hydronium ionto pH, spaced logarithmically from 0.001 to 0.1 mol/liter with 10 values. Assuming that you havenamed the concentration of hydronium ions H_conc, the syntax for calculating the negative of thelogarithm of the concentration (and thus the pH) is:pH = -log10 (H_conc)
If an Ecell of 0.495V was noted when your cell was set-up. What is thesolution pH?
In the standardi zation of HCl using pure anhydrou s sodium carbonate as primarystandard for methy l orange as indica tor , 1.0 mL HCl was found to be equiva lent to 0.05gof sodium carbonate (MW =106). The no rmality of HCl is
Trial 1: Initial pH= 3.92mL Mass of KHP and Paper= 0.868g Mass of paper= 0.357g Mass of KHP= 0.511g Trial 2: Initial pH= 4.09ml Mass of KHP and Paper= 0.870g Mass of paper= 0.359g Mass of KHP= 0.511g 1) find Ka and use the value of Ka to calculate the initial pH of each trial 2) calculate Kb for the C8H4O4^-2 anion from the value of Ka that you obtained for KHP.
Using the experimental data at 1/2 the equivalence point volume, calculate the Ka of acetic acid. Volume of NaOH at Equivalence point for HCl Trial = 5.083ml Equivalence point pH for HCl Trial = 6.57 pH 1/2 Equivalence point for HCl Trial = 3.285 Volume of NaOH at Equivalence point for HC2H3O2 Trial= 4.025ml Equivalence point pH for HC2H3O2 Trial = 4.13 pH 1/2 Equivalence point for HC2H3O2 Trial = 2.065
6 I- + BrO3- + 6 H+ ----- 3 I2 + Br- + 3 H2O Rate= k[I-]m[BrO3-]n[H+]p rate= (∆[I2])/(3*∆time) [I-] 0.010M [S2O3-2] 0.0010M H2O [BrO3-] 0.04M HCl 0.10M Starch Indicator (sec.) Volume, ml Volume, ml Volume, ml Volume, ml Volume, ml 1 10 10 20 10 10 1-3 drops 301.90 2 20 10 10 10 10 1-3 drops 213.28 3 10 10 10 20 10 1-3 drops 214.92 4 10 10 10 10 20 1-3 drops 86.45 Temperature of Reaction: _______20 C________ 1. Using the dilution equation (M1V1 = M2V2) calculate the concentration of the I-, BrO3- and the H+ (HCl) in the final mixture just prior to any reaction. 2. Calculate the rate for each experiment as explained in the background information. 3. Using the information above determine the rate law (m, n and p) for the reaction. Show all calculations/reasoning in your notebook. m_____________ n_________________ p…