Assume that pKa is 4.76

Mix 50 mL of 0.1M $\ce{HA}$ & 50 mL 0.1M $\ce{NaA}$. This is solution (i). Calculate concentrations of $\ce{HA}$ & $\ce{A-}$, measure the $\text{pH}$.

Then, mix 10 mL of solution (i) and 90 mL of water. Calculate $\ce{[HA]}$ & $\ce{[A-]}$, and measure the $\text{pH}$.

How do I approach these calculations? I have 13 of these to do, these are the first two... Equation is: $$\text{pH}=\text{pKa}+\log{\left(\frac{\ce{[A-}} {\ce{HA}}\right)}$$

For the first one, I got, $$\ce{4.76 + \log{\left(\frac{0.1}{0.1}\right)}=4.76}$$ But I don't know if that's correct. How would I approach the subsequent problem when water is added?

Assume that $\ce{pKa}$ is 4.76

Mix $\pu{50 mL}$ of $\pu{0.1M}$ $\ce{HA}$ and $\pu{50 mL}$ $\pu{0.1M}$ $\ce{NaA}$. This is solution (i). Calculate concentrations of $\ce{HA}$ and $\ce{A-}$, measure the $\ce{pH}$.

Then, mix $\ce{10 mL}$ of solution (i) and $\ce{90 mL}$ of water. Calculate $\ce{[HA]}$ and $\ce{[A-]}$, and measure the $\ce{pH}$.

How do I approach these calculations? I have 13 of these to do, these are the first two.

Equation is: $$\ce{pH} = \ce{pKa}+\log{\left(\frac{\ce{[A-]}} {\ce{[HA]}}\right)}$$

For the first one, I got, $$4.76 + \log{\left(\frac{0.1}{0.1}\right) = 4.76}$$ But I don't know if that's correct. How would I approach the subsequent problem when water is added?

Assume that pKa is 4.76

Mix 50 mL of 0.1 M HA & 50 mL 0.1 NaA. This is solution "A". Calculate concentrations of HA & A-, measure the pH.

Then, mix 10 mL of solution "A" & 90 mL of water. Calculate HA & A-, measure the pH.

How do I approach these calculations? I have 13 of these to do, these are the first two... Equation is: pH = pKa + log(mol A-/mol HA)

For the first one, I got 4.76 + log(0.1/0.1) = 4.76, but I don't know if that's correct... and then how would I approach the subsequent problem when water is added?

Assume that pKa is 4.76

Mix 50 mL of 0.1M $\ce{HA}$ & 50 mL 0.1M $\ce{NaA}$. This is solution (i). Calculate concentrations of $\ce{HA}$ & $\ce{A-}$, measure the $\text{pH}$.

Then, mix 10 mL of solution (i) and 90 mL of water. Calculate $\ce{[HA]}$ & $\ce{[A-]}$, and measure the $\text{pH}$.

How do I approach these calculations? I have 13 of these to do, these are the first two... Equation is: $$\text{pH}=\text{pKa}+\log{\left(\frac{\ce{[A-}} {\ce{HA}}\right)}$$

For the first one, I got, $$\ce{4.76 + \log{\left(\frac{0.1}{0.1}\right)}=4.76}$$ But I don't know if that's correct. How would I approach the subsequent problem when water is added?