Given: NaHSO_{4}, NaHC_{2}O_{4}, NaH_{2}PO_{4}, and NaHCO_{3}

The cation in all of the given salts is Na^{+}: Na^{+} is a main group metal cation with a +1 charge → **Na ^{+} is a neutral ion**

Therefore, we have to look at the anion. **Compare K _{a} and K**

Recall that

$\overline{){{\mathbf{K}}}_{{\mathbf{w}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{\times}}{{\mathbf{K}}}_{{\mathbf{b}}}}$

The K_{b} of S^{2}^{–} is:

$\overline{){{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{=}}\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{a}\mathbf{2}}}}$

**For ****NaHSO**_{4}**:** NaHSO_{4} dissociates to form **Na ^{+}** and

*Anion:* HSO_{4}^{–}

K_{a2} of H_{2}SO_{4} = **1.2 × 10 ^{–2}**

The K_{b} of HSO_{4}^{–} is:

${\mathbf{K}}_{\mathbf{b}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{14}}}{\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}}$

**K _{b} = 8.33 × 10^{–2}**

**K _{a} > **

**NaHSO _{4} **will produce an acidic solution.

**For ****NaHC _{2}O**

*Anion:* HC_{2}O_{4}^{–}

K_{a2} of H_{2}C_{2}O_{4} = **6.4 × 10 ^{–5}**

Using the table below, how many of the following salts are expected to produce acidic solutions: NaHSO_{4}, NaHC_{2}O_{4}, NaH_{2}PO_{4}, and NaHCO_{3}?

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