How to prepare a buffer in lab?
We are doing a lab in my chemistry class where we are assigned a pH and we have to choose from given reagents which two will be the best to make our buffer. My ph = 10.6 and I have chosen my 2 reagents with a pka closest to that. What exactly is the next step? In my lab it says "Your buffer should be .05M in the conjugate acid, and 100 mL total volume. If both of your reagents are available as solid, you should calculate the mass of each reagents you are going to mix. If only 1 reagent is available as a solid, you will have to add .50M HCl to NaOH to create its conjugate form in solution. Show calculations please.
How to prepare a buffer in lab?
We are doing a lab in my chemistry class where we are assigned a pH and we have to choose from given reagents which two will be the best to make our buffer. My ph = 10.6 and I have chosen my 2 reagents with a pka closest to that. What exactly is the next step? In my lab it says "Your buffer should be .05M in the conjugate acid, and 100 mL total volume. If both of your reagents are available as solid, you should calculate the mass of each reagents you are going to mix. If only 1 reagent is available as a solid, you will have to add .50M HCl to NaOH to create its conjugate form in solution. Show calculations please.
Suppose you are in the lab and need to prepare a CH3COOH/CH3COO- buffer of pH=5.1. Ka of CH3COOH=1.80x10^-5.?
Outline a plan to do this under the following restrictions: A. You must prepare exactly 100.0mL of the buffer solution. B. You will use 1.00 M acetic acid as the source of acetic acid AND you will make the buffer solution 0.01 M in acetic acid (you will have to determine the volume of 1M acetic acid needed). C. You will need to calculate the mass of sodium acetate to use.
Suppose you are in the lab and need to prepare a CH3COOH/CH3COO- buffer of pH=5.1. Ka of CH3COOH=1.80x10^-5.?
Outline a plan to do this under the following restrictions: A. You must prepare exactly 100.0mL of the buffer solution. B. You will use 1.00 M acetic acid as the source of acetic acid AND you will make the buffer solution 0.01 M in acetic acid (you will have to determine the volume of 1M acetic acid needed). C. You will need to calculate the mass of sodium acetate to use.
What is the pH of a mixture of 0.05M of CH3COOH and 0.1M CH3COO?
I don't answer numerical questions with numerical answers because then it feels like I'm doing someone's homework. But I'll put you on the right track.You have the concentrations of both acetic acid and acetate so you can work out the Ka of the system. Once you have that you can use an equation to get the pKa of the system. Finally you can work out the pH from there.Hope that helps!
A buffer was prepared by adding acetic acid and sodium acetate, write an equation for the reaction that occurs when a few drops of HCl are added 1?
Acetic acid is a weak acid. Sodium hydroxide is used here, which is a strong base. SO, this is WASB buffer solution. Sodium hydroxide is completely ionized. This ionization demands more acetate ions, thus more disassociation of acetic acid (Le Chatterley's Principle). These acetate ions, when concentration of H+ ions increase upon dilution, combines with them to form acetic acid back. equations are;CH3COOH = CH3COO- + H+NaOH = Na+ + OH-HCl = H+ + Cl-netralisation of water as product.CH3COO- + H+ = CH3COOH.
How do I prepare a buffer solution having a pH 7?
In the first method, prepare a solution with an acid and its conjugate base by dissolving the acid form of the buffer in about 60% of the volume of water required to obtain the final solution volume. Then, measure the pH of the solution using a pH probe.
A buffer is prepared in which the ratio [HCO3-]/[CO3^2-] is 4.0. What is the pH of this buffer (Ka HCO3-=4.7 x 10^-11)?
The dissociation constant (Ka) of an acid is the ratio of molar concentrations[H+][A-]/[HA]=KaOn the case of your question,[H+][CO3^2-]/[HCO3-]=4.7×10^(-11) , so[H+]=([HCO3-]/[CO3^2-])×4.7×10^(-11)=4.0×4.7×10^(-11)=18.8×10^(-11) .pH=-log([H+])=-log(18.8×10^(-11))=-(-9.73)=9.7 (rounded ).Does it make sense?:If you made the buffer with a ratio of 1.0, you would have [H+]=Ka, and since Ka is between 10^(-10) and 10^(-11), the pH would be between 10 and 11. If from there you added acid (say HCl) the H+ added would turn some CO3^2- to HCO3- . If you added acid until the ratio was 4, you would get the solution in your question. It would be more acidic than originally, and the pH would be lower than originally.
What concentrations of CH3COOH and CH3COONa are needed to prepare a 0.10M buffer at pH 5.0?
I think the Ka=1.8*10^-5Using Henderson equationPH=pka+log salt/acid5=-log Ka + log salt/acid5=-log (1.8*10^-5) + log salt/acid5= 4.745 + log salt/acidLog salt/acid=5–4.745Log salt/acid=0.2553Salt/acid=antilog(0.2553)Salt/acid=1.8001Salt=1.8001*acid…………(1)But to prepare a concentration of 0.1 MSalt +acid= 0.1Salt=0.1-acid……………(2)Substituting equation 2 in to 10.1-acid=1.8001*acid0.1=1.8001acid +acid0.1=2.8001acidAcid=0.1/2.8001Acid=0.0357ButSalt +acid =0.1Salt=0.1-acidSalt=0.1–0.0357Salt=0.0643
How does a buffer have stable pH value?
Note that a buffer can only have a stable pH only when [salt]/[acid] ( in case of an acidic buffer ) or [salt]/[base] ( in case of a basic buffer ) lies between 0.1 to 10.Stability of buffer’s pH can easily be understood through its mechanism..Lets take an acidic buffer say CH3COOH + CH3COONa soln. Suppose we add H+ ion from outside——————-Now note that due to extremely high equilibrium constant of our buffer rxnCH3COO- + H+ <=====> CH3COOH K = 10^5 ( approx)the forward tendency of the reaction will be extremely high. In other words rxn can be degree of completion of the rxn will be 98- 99%. Thus we usually assume that the entire externally added Hydrogen ions ( from a strong acid ) are consumed in the buffer rxn. Hence there is no alteration in pH.NOTE : Of course no rxn proceed to 100% completion , so in actual practise there are some H+ ions left, unreacted in the buffer soln.. However their concn is so low that their presence causes pH to alter VERY SLIGHTLY. This alteration is so small (tending to zero) that it is often neglected and thus , pH of the buffer is said to be unchanged.NOTE 2: The Keq value for dissociation of acetic acid is actually 1.5× 10^(-5) . Thats the advantage of using weak components in a buffer coz when the rxn proceed in reverse its Keq value indicates 99% rxn completion, hence CONSUMING ALL THE EXTERNALLY ADDED HYDROGEN IONS FROM AN ACID.______________ HOPE I HELPED __________________