100 mL of a buffer solution contains 0.1 M each of weak acid HA and salt NaA. How many gram of NaOH should be added to the buffer so that it pH will be 6?(Ka of HA = 10-5)

For acidic buffer, pH=pKa+log⁡0.10.1pH=pKa=−log⁡10−5=5Rule: ABA (In acidic buffer (A), on addition of SB (B), the concentration of WA(A) decreases and that of salt increases.Let x M of NaOH is added.pHnew =5+log⁡0.1+x0.1−x6−5=log⁡0.1+x0.1−x0.1+x0.1−x= Antilog (1)=10Solve for x:x=0.082M=0.0821000×100=0.0082mol(100mL)−1=0.0082×40g(100mL)−1=0.328g