When 700ml of water is added in a buffer solution containing 0.01M of CH3COOH and 0.1M CH3COONa then the pH of final solution is(Ka of CH3COOH=1.8X10^-5,log 1.8=0.25)

When 700ml of water is added in a buffer solution containing 0.01M of CH3COOH and 0.1M CH3COONa then the pH of final solution is(Ka of CH3COOH=1.8X10^-5,log 1.8=0.25)

Solution :
Step 1: Addition of 700 ml of water does not affect pH because
Step 1:The Henderson-Hasselbalch equation for the pHpH of a buffer solution of the monoprotic acid HAHA is given by
pH=pKa+log[A−][HA]
pH=pKa+log⁡[AX−][HA]
Since concentration appears in both the numerator and denominator of the fraction [A−][HA][AX−][HA] and pKapKa is constant (at a fixed temperature), it appears that dilution of the solution with pure H2OHX2O would not change the pHpH. However, since
pH=−log[H+]
pH=−log⁡[HX+]
the moles of H+HX+ must increase in order for pHpH to stay constant upon dilution.
Where is this additional H+HX+ coming from? I know that diluting an acid causes it to dissociate to a greater extent. But at the same time, you would be diluting its conjugate base and causing it to associate more, cancelling the dissociation of the acid

Step 2:pH=pKa+log10([A−][HA])pH=pKa+log10⁡([A−][HA])

where  pKa=−log10KapKa=−log10⁡Ka
1.[CH3COOH]=0.01M,[CH3COONa]=0.1M and Pka=4.75 and log 10[ CHCOO-]/[CH3COOH]=1
So,Ph=5.75(Answer )