3)
pKb of NH3 = 4.74
moles of NH3 = 50.0 x 1.0 / 1000 = 0.050 mol
pH = 9.00
pOH = 6.00
pOH = pKb + log [NH4Cl / NH3]
6.00 = 4.74 + log [NH4Cl / 0.050]
moles of NH4Cl = 0.9098
0.50 x V = 0.9098
V = 1.82
volume of NH4Cl = 1.82 L
4)
mmoles of NaOH = 10 x 0.24 = 2.4
mmoles of CH3COOH = 50 x 0.20 = 10
CH3COOH + NaOH - ----------> CH3COO- + H2O
10 2.4 0 0
7.6 0 2.4
pH = pKa + log [CH3COO- / CH3COOH]
= 4.74 + log [2.4 / 7.6]
pH = 4.24
answer question 3and 4 Question 2 (2 points) If 0.25 g of Na2HPO4 (disodium phosphate) is...
Henderson-Hasselbach equation: pH- pKa log (IA-|/IHA]) 1. Phosphate buffer is a mixture of KH2PO4 and K2HPO4. Note that KH2PO4 has one additional proton. The pKa of the acid is 6.8. Use the Henderson-Hasselbach equation (above) to calculate the ratio of [K2HPO41[KH2PO4] needed to make a solution that is pH 7.2 2. To make a solution that is 0.2 M phosphate, the concentration of KH2PO4 and K2HPO4 must add up to 0.2 M. Use the ratio you calculated above, and the...
a) What are the concentrations of Na2HPO4 and NaH2PO4 in a 0.30 M phosphate buffer solution pH 7.0? Use pKa 6.82 b) Describe how you would prepare 250 mL of the buffer in part (a) given that you have available to you a 1.0 M stock solution of NaH2PO4 and solid Na2HPO4(FW 141.96 g/mol). Provide your answer in milliliters of NaH2PO4 and grams of Na2HPO4. c) Suppose you use 100 ml of this buffer in an experiment and 0.003 mol...
17. A phosphate-based buffer was prepared by dissolving 28.4 g of Na2HPO4 and 12.0 g of NaH2PO4 in one liter of solution. a) Calculate buffer pH. b) Calculate the pH of the solution by adding 5.0 mmol of HCl to 100 mL of the buffer.
Preparation of Phosphate Buffer Rxn: Purpose: The purpose of lab this week is to prepare a 0.05M sodium phosphate buffer, use a pH meter to adjust the pH of this buffer, and to calculate theoretical pH changes upon addition of acid/ base. Your theory will then be correlated against your actual observational pH changes. Solutions to be made Molecular Weight Table Solution Volume 1.0M HCL 10ML 1.0 M NaOH 20ml 0.05M Sodium Phosphate: *?g NaH2PO4 H2O + *?g Na2HPO4 7H2O,...
1) Solution Components A 1 mL 100 mM NaH2PO4 + 9 mL 100 mM Na2HPO4 B 5 mL 100 mM NaH2PO4 + 5 mL 100 mM Na2HPO4 C 9 mL 100 mM NaH2PO4 + 1 mL 100 mM Na2HPO4 D 10 mL 100 mM NaH2PO4 stock solution E 10 mL 100 mM Na2HPO4 stock solution F 10 mL distilled water pKa of phosphate: 6.8 How would you calculate the pH of each equation using the Henderson Hasselbalch equations? 2) Tris...
question 2- all 4 parts. please type orwritr clearly thank
you!
2. Buffer capacity refers to the amount of acid or base a buffer can "absorb without a significant pH change. It is governed by the concentrations of the conjugate acid and base components of the buffer. A 0.5 M buffer can "absorb" five times as much acid or base as a 0.1 M buffer for a given pH change. In this problem you begin with a buffer of known...
In our experiment, we will be using a portion of the phosphate buffer system that is based upon the following equilibrium: H2PO4- HPO42- + H+ pKa = 7.2 In this case, H2PO4- will act as the acid and HPO42- will act as the base. Materials: 1M NaOH: 40.01 g/L of solution 1M HCl: 83 mL conc. HCl/L of solution Potassium phosphate, dibasic, K2HPO4, MW= 174.18 Potassium phosphate, monobasic, KH2PO4 MW= 136.09 **I already preformed this lab, but I struggled a...
What concentrations of acetic acid (pKa 4.76) and acetate would be required to prepare a 0.10 M buffer solution at pH 4.5? Note that the concentration and/or pH value may differ from that in the first question. STRATEGY 1. Rearrange the Henderson-Hasselbalch equation to solve for the ratio of base (acetate) to acid (acetic 2. Use the mole fraction of acetate to calculate the concentration of acetate. 3. Calculate the concentration of acetic acid Step 1: Rearrange the Henderson Hasselbalch...
Calculate the pH expected for the 0.0100 M HCl solution
used in part A.
I'm so lost please help
Intermediate value Final value 1a. Calculate the pH expected for the 0.0100 M HCl solution used in part A. 1b. Calculate the percent error between the expected pH and your measured pH of the 0.0100 M HCI. our To ☺ QUESTION 2 Intermediate value Final value 2a. Calculate the pH of 0.100 M CH3COOH (approximation method). Kg = 1.8 x 10-5...
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