Use the modified H-H shown below to calculate the blood plasma pH under the following conditions: [HCO3-] = 0.022 M and pCO₂ = 46 mmHg. Provide your answer to two decimal places. Do not enter units. Dissociation Constant pH = 6.1 + log Bicarbonate (mmol/L) HCO, 0.03 x PCO₂ CO, Solubility Constant CO₂ Partial Pressure (mmHg)

Use the modified H-H shown below to calculate the blood plasma pH under the following conditions: [HCO3-] = 0.022 M and pCO₂ = 46 mmHg. Provide your answer to two decimal places. Do not enter units. Dissociation Constant pH = 6.1 + log Bicarbonate (mmol/L) HCO, 0.03 x PCO₂ CO, Solubility Constant CO₂ Partial Pressure (mmHg)

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter14: Acids And Bases

Section14.8: Acid-base Reactions Of Salts

Problem 14.21CE

See similar textbooks

Similar questions

Amino acids are an important group of compounds. At low pH, both the carboxylic acid group (CO2H) and the amine group (NHR) are protonated. However, as the pH of the solution increases (say, by adding base), the carboxylic acid proton is removed, usually at a pH between 2 and 3. In a middle range of pHs, therefore, the amine group is protonated, but the carboxylic acid group has lost the proton. (This is called a zwitterion.) At more basic pH values, the amine proton is dissociated. What is the pH of a 0.20 M solution of alanine hydrochloride, [NH3CHCH3CO2H]Cl?
using the equation below and the principle of le chatelier explain how blood pH would be influenced if these conditions took place: CO2 + H2O ⇄ H2CO3 ⇄ CO3- + H+ please include a thorough explanation. make sure to also include what would happen with the [O2] and [CO2] in the lungs. - holding your breath for a long period of time - hyperventilation - too much intake of baking soda which has the bicarbonate ion
Calculate the concentration of bicarbonate ion, HCO3-1, in a 0.010 M H2CO3 solution that has the stepwise dissociation constants Ka1 = 4.3 × 10-7 and Ka2 = 5.6 × 10-11.
2. Phoebe was tasked to ensure that an enzyme responsible for keeping her species from extinction remains biologically active. To do so, she needs to store it in a buffer solution with a pH of 10.00. Unfortunately, only three buffer systems are available in their laboratory: Buffer System Weak Acid Component Ka Conjugate Base Salt PAH Benzoic acid, C6H5COOH 6.25 x 10-5 NaC6H5COO SAD Boric acid, H3BO3 5.81 x 107 NaH₂BO3 DOH Hydrogen cyanide, HCN 6.20 X 10-10 NaCN A. In what buffer system (PAH, SAD, or DOH) will Phoebe store the enzyme? In your solution sheet, show calculations and briefly explain to support your answer. B. Write the equilibrium reaction involved in the buffer system. Write your answer on your solution sheet C. Calculate the amount (in mL) of the 2.00 M stock solution (pH 10.00) of the same buffer system needed to prepare 250 mL of a 0.650 M buffer pH 10.00.
BUFFERS, HYDROLYSIS OF SALTS 1. A 0.0250 M Ca(X)2 has a pH of 8.049. Ca(X)2 is composed of a Ca2+ cation and an unknown anion X, which undergoes hydrolysis. A. Write the balanced dissociation reaction of Ca(X)2 into its ions. Write your answer on your solution sheet B. Write the balanced hydrolysis reaction of the unknown anion, X. Write your answer on your solution sheet C. Calculate the K₁ or Kb of X. D. Given the following Ką values of different weak acids, determine the identity of X. Weak acids Ka HIO 2.00 x 10-11 HCN 6.17 x 10-10 CH3COOH 1.80 x 10-5 HNO₂ 4.00 x 10-4 ts
Use the mean value of 540 for Keq to calculate the SCN− concentration in a solution whose initial Fe3+ concentration was 4.10×10−02 M and initial SCN− concentration was 1.50×10−03 M . Fe3++SCN−⇌FeNCS2+
Benzoic acid, C6H5COOH (pKa = 4.19), is only slightly soluble in water, but its sodium salt, C6H5COO Na", is quite soluble in water. In which of the following solutions will benzoic acid dissolve more readily than in pure water? O NAHCO; O LIOH O H;SO4 O HCI OK½CO3
Write the values of Y and X. H2SO4 CH3CH=CH2 + H2O CH;CH-CH2 H HỌ 2-propanol an alcohol The pKa of a protonated alcohol is about -2.5, and the pKa of an alcohol is about 15. Therefore, as long as the pH of the solution is greater than Y and less than X, more than 50% of 2-propanol will be in its neutral, nonprotonated form. Express your answers using two significant figures separated by a comma. AX中 Y, X = %3|
The 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.
Given the chemical equation for the anthocyanin extract of red cabbage as an indicator. HIn + H20 S H;0* + In¯ Using the results from the experiment, identify which of the reactant/product are responsible for the red and green color of the indicator. Notes: solution turns red after adding acid; solution turns blue/green after adding base RED: BLUE/GREEN: Why is the color violet/blue at pH near 7? Answer in 1 sentence. Hint: What is the relationship of the species responsible for the observed colors?
Calculate the ratio of ammonium chloride to ammonia that is required to make a buffer solution with a pH of 9.00 given that Ka = [H3O+][NH3] = 5.6 ×× 10–10 for ammonium ion. NH4+(aq) + H2O(l) ⇌⇌ H3O+(aq) + NH3(aq) Give your answer to 2 decimal places.
5. a) The Kw value that we normally use is for room temperature (25°C). Calculate the equilibrium H+ concentration and pH of pure, neutral water at 0.0°C (Kw= 1.14 x 10-15at 0.0°C) b) Methylamine (CH3NH2) is a derivative of the weak base ammonia. Write the equation for the ionization of methylamine in water, and use the I.C.E. method to find the pH of a 3.75 M CH3NH2 solution.(Kb= 4.37 x 10-4at 25°C)