What will be the change in pH by adding 0.1 | vedclass

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(N/A) $1$. Calculate the $pH$ of $0.1 \ M \ NH_4OH$ (weak base): $[OH^-] = \sqrt{K_b 	imes C} = \sqrt{1.77 	imes 10^{-5} 	imes 0.1} = \sqrt{1.77

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(N/A) $1$. Calculate the $pH$ of $0.1 \ M \ NH_4OH$ (weak base):
$[OH^-] = \sqrt{K_b \times C} = \sqrt{1.77 \times 10^{-5} \times 0.1} = \sqrt{1.77 \times 10^{-6}} \approx 1.33 \times 10^{-3} \ M$.
$pOH = -\log(1.33 \times 10^{-3}) \approx 2.876$.
$pH = 14 - 2.876 = 11.124$.
$2$. Calculate the $pH$ of the buffer solution formed by adding $0.1 \ M \ NH_4Cl$ to $0.1 \ M \ NH_4OH$:
Using Henderson-Hasselbalch equation: $pOH = pK_b + \log(\frac{[Salt]}{[Base]})$.
$pK_b = -\log(1.77 \times 10^{-5}) \approx 4.752$.
$pOH = 4.752 + \log(\frac{0.1}{0.1}) = 4.752 + 0 = 4.752$.
$pH = 14 - 4.752 = 9.248$.
$3$. Change in $pH = 11.124 - 9.248 = 1.876$.

What will be the change in $pH$ by adding $0.1 \ M \ NH_4Cl$ to $0.1 \ M \ NH_4OH$ (weak base) solution? (Given: $K_b$ of $NH_4OH = 1.77 \times 10^{-5}$)

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