Which composition will make the basic buffer?

Which one of the following pairs of solutions is not an acidic buffer?

1. ${\mathrm{HClO}}_4$ $\mathrm{and}$ ${\mathrm{NaClO}}_4$
2. ${\mathrm{CH}}_3\mathrm{COOH}$ $\mathrm{and}$ ${\mathrm{CH}}_3\mathrm{COONa}$
3. ${\mathrm{H}}_2{\mathrm{CO}}_3$ $\mathrm{and}$ ${\mathrm{Na}}_2{\mathrm{CO}}_3$
4. ${\mathrm{H}}_3{\mathrm{PO}}_4$ $\mathrm{and}$ ${\mathrm{Na}}_3{\mathrm{PO}}_4$

Buffer solutions have constant acidity and alkalinity because:
 

A buffer solution is prepared in which the concentration of NH₃ is 0.30 M and the concentration of  $N{H}_4^{+\hspace{0.17em}}$ is 0.20 M. If the equilibrium constant, K_b for NH₃ equals 1.8×10^–5, then what is the pH of this solution? 
(log 1.8 = 0.25; log 0.67 = –0.176)

1.  9.43
2.  11.72
3.  8.73
4.  9.08

Calculate the hydrogen ion concentration, [\(\text{H}^+\)] (in \(\text{mol}/\text{L}\)), of a buffer solution prepared by mixing \(0.10 \text{ M}\) acetic acid (\(\text{CH}_3\text{COOH}\)) and \(0.20 \text{ M}\) sodium acetate (\(\text{CH}_3\text{COONa}\)), given that the acid dissociation constant (\(\text{K}_a\)) for acetic acid is \(1.8 \times 10^{-5}\):

1. \(3 . 5 \times 10^{- 4}\)
2. \(1 . 1 \times 10^{- 5}\)
3. \(1 . 8 \times 10^{- 5}\)
4. \(9 . 0 \times10^{- 6}\)${\mathrm{}}^{}$

In a buffer solution containing an equal concentration of B^- and HB, the K_b for B^- is 10^-10. pH of the buffer solution is:

The following pair constitutes a buffer is:

1. $HN{O}_2$ $and$ $NaN{O}_2$

2. $NaOH$ $and$ $NaCl$

3. $HN{O}_3$ $and$ $N{H}_{4}N{O}_3$

4. $HCl$ $and$ $KCl$

The rapid change of pH near the stoichiometric point of an acid-base titration is the basis of indicator detection. pH of the solution is related to the ratio of the concentrations of the conjugate acid (\(HIn\)) and base (\(In^–\)) forms of the indicator, as per the expression:

1. $\log \frac{\left[\mathrm{HIn}\right]}{\left[{\mathrm{In}}^{-}\right]}={\mathrm{pK}}_{\mathrm{In}}-\mathrm{pH}$

2. $\log \frac{\left[\mathrm{HIn}\right]}{\left[{\mathrm{In}}^{-}\right]}=\mathrm{pH}-{\mathrm{pK}}_{\mathrm{In}}$

3. $\log \frac{\left[{\mathrm{In}}^{-}\right]}{\left[\mathrm{HIn}\right]}=\; pH\; +\hspace{0.17em}{\mathrm{pK}}_{\mathrm{In<mspace\; width="1em"></mspace>}}$

4. None of the above