Question

match the following ph values: 1, 2, 5, 6, 6.5, 8, 11, 11, and 13 with the following chemicals (of equal concentration): hbr, naoh, naf, nacn, nh4f, ch3nh3f, hf, hcn, and nh3. answer this question without performing calculations. hbr □ naoh □ naf □ nacn □ nh4f □ ch3nh3f □ hf □ hcn □ nh3 □

Explanation:

• Strong acid: \(HBr\) is a strong acid. Strong acids completely dissociate in water, so a solution of \(HBr\) will have a very low \(pH\). Among the given \(pH\) values, \(pH = 1\) is the most acidic.
• Strong base: \(NaOH\) is a strong base. Strong bases completely dissociate in water, so a solution of \(NaOH\) will have a very high \(pH\). Among the given \(pH\) values, \(pH=13\) is the most basic.
• Weak acid - conjugate base (salt of weak acid and strong base):
• For \(NaF\), \(F^{-}\) is the conjugate base of \(HF\) (\(HF

ightleftharpoons H^{+}+F^{-}\), \(K_{a}(HF)=6.6\times 10^{-4}\)). The salt \(NaF\) hydrolyzes as \(F^{-}+H_{2}O
ightleftharpoons HF + OH^{-}\). Since \(HF\) is a weak acid, \(NaF\) solutions are basic. But it is a weaker base compared to \(NaCN\) (because \(HCN\), \(K_{a}(HCN)=6.2\times 10^{-10}\) is a weaker acid than \(HF\)). So \(NaF\) has a \(pH = 8\)

• For \(NaCN\), \(CN^{-}\) is the conjugate base of \(HCN\). Since \(HCN\) is a very weak acid (\(K_{a}(HCN)=6.2\times 10^{-10}\)), \(CN^{-}\) hydrolyzes more than \(F^{-}\) (\(CN^{-}+H_{2}O

ightleftharpoons HCN+OH^{-}\)). So \(NaCN\) has a higher \(pH\) than \(NaF\). \(NaCN\) has \(pH = 11\)

• Weak acid - weak base salts:
• For \(NH_{4}F\), \(NH_{4}^{+}\) is the conjugate acid of \(NH_{3}\) (\(NH_{3}+H_{2}O

ightleftharpoons NH_{4}^{+}+OH^{-}\), \(K_{b}(NH_{3}) = 1.8\times 10^{-5}\), so \(K_{a}(NH_{4}^{+})=\frac{K_{w}}{K_{b}}=\frac{10^{-14}}{1.8\times 10^{-5}}\approx5.6\times 10^{-10}\)) and \(F^{-}\) is the conjugate base of \(HF\) (\(K_{a}(HF) = 6.6\times 10^{-4}\)). Since \(K_{a}(HF)>K_{a}(NH_{4}^{+})\), the solution of \(NH_{4}F\) is slightly basic (\(pH = 6.5\))

• For \(CH_{3}NH_{3}F\), \(CH_{3}NH_{3}^{+}\) is the conjugate acid of \(CH_{3}NH_{2}\) (\(K_{b}(CH_{3}NH_{2})=4.4\times 10^{-4}\), so \(K_{a}(CH_{3}NH_{3}^{+})=\frac{K_{w}}{K_{b}}=\frac{10^{-14}}{4.4\times 10^{-4}}\approx2.3\times 10^{-11}\)) and \(F^{-}\) is the conjugate base of \(HF\) (\(K_{a}(HF) = 6.6\times 10^{-4}\)). Since \(K_{a}(HF)>K_{a}(CH_{3}NH_{3}^{+})\), but the difference is larger than in \(NH_{4}F\), \(CH_{3}NH_{3}F\) is more basic than \(NH_{4}F\) but less basic than \(NaF\). However, considering the relative \(K\) values and the fact that \(HF\) is a stronger acid than \(CH_{3}NH_{3}^{+}\), \(CH_{3}NH_{3}F\) has \(pH = 6\)
• Weak acids:
• For \(HF\), \(K_{a}(HF)=6.6\times 10^{-4}\). Among the weak acids (\(HF\) and \(HCN\)), \(HF\) is a stronger acid. So \(HF\) has \(pH = 2\)
• For \(HCN\), \(K_{a}(HCN)=6.2\times 10^{-10}\). Since \(HCN\) is a weaker acid than \(HF\), \(HCN\) has \(pH = 5\)
• Weak base: \(NH_{3}\) is a weak base (\(NH_{3}+H_{2}O

ightleftharpoons NH_{4}^{+}+OH^{-}\), \(K_{b}(NH_{3})=1.8\times 10^{-5}\)). It is a weaker base than \(NaCN\) and \(NaOH\). So \(NH_{3}\) has \(pH = 11\) (but note that \(NaCN\) and \(NH_{3}\) have the same \(pH\) value here, but \(NaOH\) is a strong base, \(NaCN\) is a salt - based base (from hydrolysis of \(CN^{-}\)) and \(NH_{3}\) is a molecular weak base)

Answer:

• \(HBr:1\)
• \(NaOH:13\)
• \(NaF:8\)
• \(NaCN:11\)
• \(NH_{4}F:6.5\)
• \(CH_{3}NH_{3}F:6\)
• \(HF:2\)
• \(HCN:5\)
• \(NH_{3}:11\)