Calculate the molar concentration of ethanoic acid needed to prepare ethanoate buffer of PH 6.8 using 0.5m solution of sodium ethanoate. Pka of ethanoate acid is 0.127
Do you mean 0.5M. M means molar. m means molal.
pH = pKa + log (base)/(acid)
6.8 = 0.127 + log (0.5M/x)
Sovle for x.
To calculate the molar concentration of ethanoic acid needed to prepare an ethanoate buffer of pH 6.8 using a 0.5 M solution of sodium ethanoate, we can utilize the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa and the ratio of the concentrations of the conjugate acid and its conjugate base.
The Henderson-Hasselbalch equation is given as:
pH = pKa + log([A-]/[HA])
Where:
pH is the desired pH of the buffer (6.8)
pKa is the dissociation constant of the acid (0.127)
[A-] is the concentration of the conjugate base (sodium ethanoate)
[HA] is the concentration of the acid (ethanoic acid)
First, we can rearrange the Henderson-Hasselbalch equation to solve for [HA]:
pH - pKa = log([A-]/[HA])
Next, we apply the antilog function (10^x) to both sides of the equation to remove the logarithm:
10^(pH - pKa) = ([A-]/[HA])
Now, let's substitute the given values into the equation:
10^(6.8 - 0.127) = ([A-]/[HA])
Solving this equation will give us the ratio of [A-] to [HA]. Since the concentration of sodium ethanoate is given as 0.5 M, we can substitute this into the equation:
10^(6.8 - 0.127) = (0.5 M/[HA])
Simplifying further:
10^6.673 = (0.5 M/[HA])
Which gives us:
398744.1135 = (0.5 M/[HA])
Now we can solve for [HA]:
[HA] = 0.5 M / 398744.1135
Calculating this value will give us the molar concentration of ethanoic acid needed to prepare the ethanoate buffer of pH 6.8 using the 0.5 M solution of sodium ethanoate.