Question

A 1.0l buffer solution contains 0.100 mol of hc2h3o2 and 0.100 mol of nac2h3o2. the value of ka for hc2h3o2 is 1.8×10−5. part a calculate the ph of the solution upon the addition of 0.015 mol of naoh to the original buffer. express the ph to two decimal places.

Answer 1

Hey there!:

Ka = 1.8*10⁻⁵

Pka = 4.74

add the moles of  NaOH to moles of base:

0.015 moles + 0.100 moles => 0.115 moles

Since you add mol base you have to take away that same , amount from acid:

0.100 acid  - 0.015  NaOH = 0.085

Therefore :

pH = pKa + log [ A ] / [ HA ]

pH = 4.74 + log [ 0.115 ] / [ 0.085 ]

pH =  4.87600

Hope that helps!

Answer 2

The estimation of the acidic and basic character of the solution by the hydrogen and hydroxide ion is called pH. The pH of the solution is 4.876.

What is pH?

pH is the measure of the hydrogen or the hydroxide ions released in the solution that depicts the acidity or the basicity of the solution and is measured by the pH scale.

Given,

• Ka = $1.8 \times 10^{-5}$
• Moles of NaOH = 0.015 mmol
• Initial moles of $\rm C_{2}H_{3}O_{2}$ = 0.1 mmol
• Initial moles of $\rm HC_{2}H_{3}O_{2}$ = 0.1 mmol

After reaction the final moles of $\rm C_{2}H_{3}O_{2}$ :

$\begin{aligned} \rm Initial\; moles + Added \; moles &= 0.1 + 0.015\\\\&= 0.115 \;\rm mmol\end{aligned}$

After reaction the final moles of $\rm HC_{2}H_{3}O_{2}$:

$\begin{aligned} \rm Initial \;moles - Added \; moles &= 0.1 - 0.015\\\\&= 0.085 \;\rm mmol\end{aligned}$

Calculate pKa:

$\begin{aligned} \rm pKa &= \rm -log (Ka)\\\\&=\rm -log (1.8 \times 10^{-5})\\\\&= 4.74\end{aligned}$

Using the formula for pH:

$\begin{aligned} \rm pH &= \rm pKa + log \dfrac{[A]}{[HA]}\\\\&= 4.74 +\rm log \dfrac {0.115}{0.085}\\\\&= 4.87\end{aligned}$

Therefore, the pH of the solution is 4.87.

Learn more about pH here:

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