Topic: pH and Ka Value

Note:
y-intercept = 8.46
Henderson-Hasselbalch equation : pH = pKa + log[A-/HA]
Literature value of pKa = 9.245
Question: Use Henderson-Hasselbalch equation and y-intercept to determine the Ka value for the acid in the conjugate pair (NH3 and NH4Cl). Show your work.
Answer – We are given the Henderson-Hasselbalch equation and graph between the pH vrs log [NH3]/[NH4Cl]
We know the Henderson-Hasselbalch equation-
pH = pKa + log[A-/HA]
From the graph , y-intercept = 8.46
When we draw the graph between the pH vrs log salt/acid then there is straight line equation and the intercept is the pKa values.
So, y-intercept = pKa = 8.46 when graph between the pH vrs log salt/acid.
We know,
pKa = - log Ka
so, Ka = 10-pKa
= 10-8.46
= 3.46*10-9
The actual Ka from pKa = 9.245 is 5.68*10-10, so there is maybe some error in the calculation or recording data.
Topic: pH and Ka Value Note: y-intercept = 8.46 Henderson-Hasselbalch equation : pH = pKa +...
The pH of a buffer is calculated by using the Henderson-Hasselbalch equation: pH=pKa +log[Base]/[Acid] Part A: What is the pH of a buffer prepared by adding 0.809mol of the weak acid HA to 0.406mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66
1)Which expression is the correct form of the Henderson-Hasselbalch equation? pH = pKa + log([base]/[acid]) pKa = pH + log([base]/[acid]) pH = pKa − log([base]/[acid]) pH = pKa + log([acid]/[base]) 2) Use the Henderson-Hasselbalch equation to calculate the pH of a buffer solution prepared by mixing equal volumes of 0.220 M NaHCO3 and 9.00×10−2 M Na2CO3. (Ka values are given in Appendix C.) Express your answer using three significant figures.
Henderson-Hasselbalch equation explain the variables and constants. pH=pKa+log[A-]/[HA]
% Ionization using Henderson Hasselbalch equation ionization given ph and pka: pka:8.0 at a pH of 7.4 what is the percent ionization? Here is what I have so far: pka=pH + log(acid/base) 8.0=7.4+ log (acid/base) 0.6= log (acid/base) 10^0.6= (acid/base) = 3.98 this is where I got stuck. my solution should be 80% but I not sure how. please show all work. Book answer: % acid form = (3.98 x100)/ 4.98 = 79.99 % Where did they get 4.98???
The Henderson-Hasselbalch equation connects pH to pk, by relating pH to the relative amounts of the acid and conjugate base. The equation is: [A], pH = pKa + log [HA]' A. If you had an acetic acid solution at pH 4.75, what would the ratio of acetic acid to acetate 4. be? (Сн,соо у сн, соон) - ([CH3CO0¯], [CH3COOH], B. What if the solution pH was 4.27? C. What about pH 5.05?
Need help on questions 1-3
Henderson-Hasselbalch: pH=pka + Log( [base]/(acid] ) 1. The value of Ka of nitrous acid (HNO2) is 4.6 x 104 M. Calculate the pH and the concentration of [H3O+] in a 0.02M aqueous solution of HNO2(aq). 2. Label conjugate acid-base pairs: HNO2(aq) + H2O → H30* + NO2 (aq) 3. What will happen to the reaction equilibrium if we increase the pressure in the reaction vessel? H2(g) + 12(e) → 2 HI(g)
QUESTION 5 (A ) pH = pka + log THA) Use the Henderson-Hasselbalch equation to solve the following problem: A solution of hypoiodous acid (HIO, pka = 10.50) has a measured pH of 11.24. What is the % ionization of the hypoiodous acid? 58.71% 15.40% 84.60% O 73.44%
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of its conjugate acid and the ratio of the concentrations of the conjugate base and acid. The equation is important in laboratory work that makes use of buffered solutions, in industrial processes where pH needs to be controlled, and in medicine, where understanding the Henderson-Hasselbalch equation is critical for the control of blood pH. Part A As a technician in a large pharmaceutical research firm, you need...
Write the Henderson-Hasselbalch equation for a solution of propanoic acid (CH3CH2CO2H, pka = 4.874) using HA, A-, and the given Pka value in the expression. Using this equation, calculate the quotient (A]/[HA] at A) pH 4.23 B) pH 4.874 C) pH 530.
please show all work
12. Using the Henderson-Hasselbalch equation: [Α] pH = pka + log Calculate what relative amounts of sodium dihydrogen phosphate and sodium monohydrogen phosphate are required to make a buffer solution with pH = 7.9.