Question

question 5 (6 points)
a buffer solution is made from 0.650 mol/l ammonia and 0.350 mol/l ammonium chloride (to total 1l of liquid). what is the ph of this buffer?
$(k_b = 1.8\times10^{-5})$
$ph = 3.4$
$ph = 8.5$
$ph = 9.5$
$ph = 4.5$

Explanation:

Step1: Calculate pKb

The formula for pKb is $pK_b = -\log(K_b)$. Given $K_b = 1.8\times10^{-5}$, then $pK_b=-\log(1.8\times 10^{-5})\approx4.74$.

Step2: Use Henderson - Hasselbalch equation for bases

The Henderson - Hasselbalch equation for a basic buffer is $pOH = pK_b+\log\frac{[salt]}{[base]}$. Here, the base is ammonia ($[base]=0.650$ mol/L) and the salt is ammonium chloride ($[salt]=0.350$ mol/L). So $pOH = 4.74+\log\frac{0.350}{0.650}=4.74+\log(0.538)\approx4.74 - 0.27 = 4.47$.

Step3: Calculate pH

Since $pH + pOH=14$, then $pH = 14 - pOH$. Substituting the value of pOH, we get $pH=14 - 4.47 = 9.53\approx9.5$.

Answer:

pH = 9.5