Consider the reaction of a 20.0 mL of 0.220 M CsHsNHCI (Ka = 5.9 x 10-) with 12.0 mL of 0.215 M CSOH. a) Write the net ionic equation for the reaction that takes place. Be sure to include the proper phases for all species within the reaction. b) What quantity in moles of CsHsNH+ would be present at the start of the titration? c) What quantity in moles of OH- would be present if 12.0 mL of OH- were added?

Fill out the tables and find pH

You have 1.5 liter of solution that is composed of 8.88 grams of NH3 and 11.33 grams of ammonium chloride mixed well. Kb for ammonia =1.8 x 10^-5 (A) is this a buffer solution? Why or why not? (B)if it is a buffer solution, what is the pH of this buffer solution? (C)how many mL of 1.50 M HCl can be added to this solution before the buffer is exhausted (d) how many mL of 1.5 M NaOH can be added to this solution before the buffer is exhausted?

5:51 1 .ll LTE Search Question 16 of 20 Submit Determine the pH at the point in the titration of 40.0 mL of 0.200 M H2NNH, with 0.100 M HNO3 after 80.0 mL of the strong acid has been added. The value of Kb for H2NNH2 is 3.0 x 10-6. 1 2 4 Based on the result of the acid-base reaction, set up the unknown. ICE table in order to determine the H2NNH3*(at H2O(1) =H;O*(aq) H2NNH2(aq + Initial (M) Change (M) Equilibrium (M) RESET 0.100 0.200 0.0333 0.0667 +x -X 0.100 + x 0.100 - x 0.200 + x 0.200 - x 0.0333 + x 0.0333 - x 0.0667 + x 0.0667 - x www

A pipette is used to add 41 mL of 0.2 M Pb(NO3)2 to a 268.0 mL solution of 0.2 M NaF. Determine the equilibrium concentration of Pb2+(aq) in the final solution. Ksp(PbF2)=3.6E-8Include at least 3 significant figures in your answer.

Answer Questions #27 through #31 about the titration curves below (labeled a and b) for two different bases. The titrant acid used in both cases was 0.100 M HCl and the volume of the base solutions in both cases was 75.0 mL mL. 14 14 12 12 - 10- 10 8- 6. 4. 2- 2- 20 40 60 80 100 20 40 60 80 100 Volume of acid added (mL) (a) (b) Volume of acid added (mL) 2014 Pearson Education Inc P Type here to search 11:17 AM (? 3/18/2021 -> ort sc 80 6. ece num ER/TY || на 0 04 2C Hd

3a) A student determines the concentration of a sodium hydroxide solution by titration with standardized KHP. S/he obtains the values: 0.190 M, 0.202 M, and 0.205 M. Should the value 0.190 M be rejected? Apply the Q Test. For three values Q must be greater than 0.94 to reject the number. Q = suspect nearest | largest - - smallest b) The student de to repeat the experiment two more The five values now include: 0.190 M, 0.202 M, 0.205 M, 0.201M and 0.203M. Use the Q Test to see if the first value may be rejected. For five values Q must be greater than 0.64 to reject the number. c) Solve for the average Molarity of the measurements from part b with and without the rejected number. Is there value in repeating an experiment several times?

A 81.0 mL sample of 0.0500 M HBrO4 is titrated with 0.100 M NaOH solution. Calculate the pH after the following volumes of base have been added. (a) 14.2 mLpH =  (b) 39.3 mLpH =  (c) 40.5 mLpH =  (d) 42.5 mLpH =  (e) 79.0 mLpH =

1. How do you prepare a 100mL of 0.1 M phosphate buffer?To make 100 mL of 0.1 M phosphate buffer: Calculate the amount of sodium phosphate needed. moles of sodium phosphate = (0.1 mol/L) x (0.1 L) = 0.01 moles mass of sodium phosphate = moles x molar mass = 0.01 moles x 142 g/mol = 1.42 g Dissolve 1.42 g of sodium phosphate in distilled water in a 100 mL volumetric flask. Adjust the pH of the solution to the desired value (usually around 7.4) using a strong acid or strong base. Bring the solution to the final volume (100 mL) with distilled water.   2. From the anterior buffer, how do you make 100mL of 0.05 M? To make 100 mL of 0.05 M phosphate buffer from the 0.1 M stock solution: Calculate the amount of the 0.1 M phosphate buffer needed. moles of phosphate buffer = (0.05 mol/L) x (0.1 L) = 0.005 moles Calculate the volume of the 0.1 M phosphate buffer needed. moles = concentration x volume (in liters) volume = moles / concentration = 0.005 moles / 0.1 mol/L = 0.05 L or 50 mL Measure…

2. Using the procedure described in this module, a student determined the percent KHP in an impure sample of KHP. A 3.150-g sample of impure KHP required 41.50 mL of 0.1352M NaOH solution for titration. (a) Calculate the number of moles of NaOH required for the titration. (b) Calculate the number of moles of KHP present in the impure sample of KHP. (c) Calculate the number of grams of KHP present in the impure sample. (d) Calculate the percent of KHP in the impure sample, using Equation 8. Equation 8: percent KHP in the impure sample, % = ( mass of KHP in the sample,g/ mass of sample analyzed, g) (100%)