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in Chemistry by (15 points)
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How much volume of 0.3 MNH4OH should be added to 60 mL of 0.1 MNH4Cl so that pH of the buffer solution is 10? Given that pKb for NH4OH = 4.75.

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6. According to Henderson - hassel blach equation for base buffer

POH - Pkb + log \(\frac{[salt]}{[base]}\) ----- (1)

we have given,

PH of buffer solution = 10

∴ POH = 14 - PH

 = 14 - 10

POH = 4

Let say volume of Hn4OH = VmL

Then number of moles of NH4OH = V x 0.3 milimol

Total volume of buffer solution = (V + 60) ml

Number of moles of NH4CL = 60 x 0.1 = 6 milimol

Putting the values in equation (1) -- we got.

taking anti log on both side---

0. 178 = \(\frac{6}{V\times0.3}\)

\(\Rightarrow{}\)v = \(\frac{6}{0.178\times0.3}\)

\(\Rightarrow{}\) v = 112.4 ml

Hence, 112.4 ml of 0.3 M NH4OH should be added.

7. According to Henderson - Hassel blach equation for acid buffer --

PH = PKA + log \(\frac{[salt]}{[acid]}\) -----(II)

we have given,

concertrational of CH3COOH = 0.01 M

concertrational of CH3COOHa = 0.4 M

volume of CH3COOH = V ML

Let see, Volume of CH3COOH = V ML

Then, Total volume of buffer solution = (V + 25) ML

PH of buffer solution = 4.91

PKA of acid = 4.76

After mixing --- 

concertrational of CH3COOH = \(\frac{V\times0.01}{V+25}\) M

concertrational of CH3COOHa = \(\frac{10}{V+25}\) M

Putting the values in equation (II) we got --

4.91 = 4.76 + log \(\frac{(\frac{10}{V+25})}{(\frac{V\times0.01}{V+25})}\) 

\(\Rightarrow\) 0.15 = log \(\frac{10}{V\times0.01}\)

taking antilog on both side --- we got ---

\(\Rightarrow\) 1.412537 = \(\frac{10}{V\times0.01}\)

\(\Rightarrow\) V = 707.946 ML

\(\Rightarrow\) V = 707.9 ML

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