a buffer is prepared by mixing .208 L of 0.452 M HCl and 0.5 L of 0.4 M sodium acetate. When the Ka= 1.8 x 10^-5
I figured out that the pH= 4.74
but then it asks how many grams of KOH must be added to the buffer to change the pH by .155 units?
Show me how you obtained pH = 4.74. I can't get that. My answer looked more like 4.79.
Since Ka a measure of H+ ions,
-log(Ka)= pH.. my assignment says that part is right, I just don't know what to do from there to get M -> moles -> grams
-log Ka = pKa which may or may not be = pH. Ka is a measure of how strong/weak the acid is.
To find out how many grams of KOH must be added to the buffer to change the pH by 0.155 units, we need to first understand how the addition of KOH will affect the pH of the buffer.
KOH is a strong base that will react with the weak acid present in the buffer (in this case, acetic acid, CH3COOH). The reaction between the two will result in the formation of water (H2O) and the conjugate base of acetic acid, acetate ion (CH3COO-).
CH3COOH + KOH → CH3COO- + H2O
The addition of KOH will cause an increase in the concentration of acetate ion in the buffer solution. This increase in the concentration of the conjugate base will shift the equilibrium of the buffer system and affect the pH.
Now, let's calculate the initial concentration of acetate ion (CH3COO-) in the buffer solution.
Initial moles of sodium acetate = volume (in liters) × molarity
= 0.5 L × 0.4 M
= 0.2 moles
Since sodium acetate is a strong electrolyte, it dissociates completely, yielding one mole of acetate ion for every mole of sodium acetate. Therefore, the initial concentration of acetate ion (CH3COO-) in the buffer solution is also 0.2 M.
Next, we need to determine the new concentration of acetate ion after the addition of KOH. To do this, we will use the formula for pH.
pH = pKa + log([Base]/[Acid])
Given that the initial pH of the buffer is 4.74, we can rearrange the equation to find the initial ratio of [Base] to [Acid].
log([Base]/[Acid]) = pH - pKa
log([Base]/0.2) = 4.74 - (-log(1.8 × 10^-5))
Now, calculate the ratio of [Base] to [Acid] by taking the antilog of both sides of the equation.
[Base]/0.2 = 10^(4.74 - (-log(1.8 × 10^-5)))
Solve for [Base].
[Base] = 0.2 × 10^(4.74 - (-log(1.8 × 10^-5)))
Now, we need to determine the new concentration of acetate ion after the pH change of 0.155 units. We can use the Henderson-Hasselbalch equation to calculate this.
pH = pKa + log([Base]/[Acid])
Using the new pH = 4.74 + 0.155, we can rearrange the equation to find the new ratio of [Base] to [Acid].
log([Base]/[Acid]) = new pH - pKa
log([Base]/[Acid]) = 4.74 + 0.155 - (-log(1.8 × 10^-5))
Now, calculate the new ratio of [Base] to [Acid] by taking the antilog of both sides of the equation.
[Base]/[Acid] = 10^(4.74 + 0.155 - (-log(1.8 × 10^-5)))
Finally, calculate the new concentration of acetate ion ([Base]) after the pH change.
[Base] = [Acid] × 10^(4.74 + 0.155 - (-log(1.8 × 10^-5)))
To determine the mass of KOH required to cause this change, we need to first find the moles of acetate ion ([Base]) needed.
Moles of acetate ion = [Base] × volume (in liters)
= [Base] × 0.5 L
Finally, we need to convert the moles of acetate ion to the mass of KOH using the molar mass of KOH (56.11 g/mol).
Mass of KOH = Moles of acetate ion × molar mass of KOH
By plugging in the values obtained from the calculations, you will be able to find the number of grams of KOH required to change the pH of the buffer solution by 0.155 units.