Answer:
0.387 g
Step-by-step explanation:
pH of the buffer = 1
V = Volume of solution = 100 mL
[HA] = Molarity of HA = 0.1 M
= Acid dissociation constant =
(assuming base as
)
Molar mass of base = 322.2 g/mol
pKa is given by
![pK_a=-\log K_a\\\Rightarrow pKa=-\log(1.2* 10^(-2))\\\Rightarrow pK_a=1.92](https://img.qammunity.org/2021/formulas/chemistry/college/uu8y8ake074htndphgm4fntkoyln530mox.png)
From the Henderson-Hasselbalch equation we get
![pH=pK_a+\log([A^-])/([HA])\\\Rightarrow pH-pK_a=\log([A^-])/([HA])\\\Rightarrow 10^(pH-pK_a)=([A^-])/([HA])\\\Rightarrow [A^-]=10^(pH-pK_a)[HA]\\\Rightarrow [A^-]=10^(1-1.92)*0.1\\\Rightarrow [A^-]=0.01202\ \text{M}](https://img.qammunity.org/2021/formulas/chemistry/college/soa3rohckpbnidokir23f0f0ir9k5vn75b.png)
Moles of base
![0.01202*100*(1)/(10^3)=0.001202\ \text{moles}](https://img.qammunity.org/2021/formulas/chemistry/college/hz7rn7a4fesnu40q92jv2xygczobrxi5s7.png)
Mass of base is given by
![0.001202* 322.2=0.387\ \text{g}](https://img.qammunity.org/2021/formulas/chemistry/college/o6aqcqec1xfp6tt2bb95r5apthgppwbv7s.png)
The required mass of the base is 0.387 g.