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Does dissolving salt in water change the pH?
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Does dissolving salt in water change the pH?
When you dissolve a salt in water, you have to distinguish 4 cases:
The salt results from the reaction of a strong acid with a strong base, the pH will be 7. This is the case, for example, of $\ce{NaCl}$ that results from the reaction of $\ce{HCl}$ and $\ce{NaOH}$. The ions $\ce{Na^+}$ and $\ce{Cl^-}$ are spectator ions and don't react with water. So, the pH doesn't change.
The salt results from the reaction of a strong acid with a weak base, the pH will be less than 7. This is the case , for example, of $\ce{NH4Cl}$ that results from the reaction of $\ce{HCl}$ and $\ce{NH3}$. The ions $\ce{Cl^-}$ are spectator ions and don't react with water. While the ions $\ce{NH_4^+}$ do hydrolyze and give ions $\ce{H^+}$ to water. So, the pH will decrease.
The salt results from the reaction of a weak acid with a strong base, the pH will be more than 7. This is the case , for example, of $\ce{CH3COONa}$ that results from the reaction of $\ce{CH3COOH}$ and $\ce{NaOH}$. The ions $\ce{Na^+}$ are spectator ions and don't react with water. While the ions $\ce{CH3COO^-}$ do hydrolyze and accept ions $\ce{H^+}$ from water. In this case, water act as an acid leaving a hydroxide ion. So, the pH will increase.
The salt results from the reaction of a weak acid with a weak base, the calculus of pH is a little bit complex. We need to take into consideration the $K_a$ and the $K_b$ of the weak acid and the weak base and whichever is the stronger acid or weak will be the dominate factor in determining whether the solution is acidic or basic. We can investigate three cases:
$K_a= K_b$, the aqueous solution is neutral.
This is the case, for example, of $\ce{NH4CH3COO}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{CH3COOH}$. The ions $\ce{NH4^+}$ and $\ce{CH3COO^-}$ do hydrolyze. As the hydrolysis constant of ion acetate equals the hydrolysis constant of ion ammonium, the same concentarion of ion hydronium and ion hydroxide is produced, and the pH of the solution is neutral.
$K_a> K_b$
This is the case, for example, of $\ce{NH4F}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{HF}$. The ions $\ce{NH4^+}$ and $\ce{F^-}$ do hydrolyze and as ammonium ions hydrolyze to a greater extent than fluoride ions, i.e more hydronium ion is produced, and the pH of the solution is acidic.
$K_a< K_b$
This is the case, for example, of $\ce{NH4CN}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{HCN}$. The ions $\ce{NH4^+}$ and $\ce{CN^-}$ do hydrolyze and as cyanide ions hydrolyze to a greater extent than fluoride ions, i.e less hydronium ions are produced, and the pH of the solution is basic.
When you dissolve a salt in water, you have to distinguish 4 cases:
The salt results from the reaction of a strong acid with a strong base, the pH will be 7. This is the case, for example, of $\ce{NaCl}$ that results from the reaction of $\ce{HCl}$ and $\ce{NaOH}$. The ions $\ce{Na^+}$ and $\ce{Cl^-}$ are spectator ions and don't react with water. So, the pH doesn't change.
The salt results from the reaction of a strong acid with a weak base, the pH will be less than 7. This is the case , for example, of $\ce{NH4Cl}$ that results from the reaction of $\ce{HCl}$ and $\ce{NH3}$. The ions $\ce{Cl^-}$ are spectator ions and don't react with water. While the ions $\ce{NH_4^+}$ do hydrolyze and give ions $\ce{H^+}$ to water. So, the pH will decrease.
The salt results from the reaction of a weak acid with a strong base, the pH will be more than 7. This is the case , for example, of $\ce{CH3COONa}$ that results from the reaction of $\ce{CH3COOH}$ and $\ce{NaOH}$. The ions $\ce{Na^+}$ are spectator ions and don't react with water. While the ions $\ce{CH3COO^-}$ do hydrolyze and accept ions $\ce{H^+}$ from water. In this case, water act as an acid leaving a hydroxide ion. So, the pH will increase.
The salt results from the reaction of a weak acid with a weak base, the calculus of pH is a little bit complex. We need to take into consideration the $K_a$ and the $K_b$ of the weak acid and the weak base and whichever is the stronger acid or weak will be the dominate factor in determining whether the solution is acidic or basic. We can investigate three cases:
$K_a= K_b$, the aqueous solution is neutral.This is the case, for example, of $\ce{NH4CH3COO}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{CH3COOH}$. The ions $\ce{NH4^+}$ and $\ce{CH3COO^-}$ do hydrolyze. As the hydrolysis constant of ion acetate equals the hydrolysis constant of ion ammonium, the same concentarion of ion hydronium and ion hydroxide is produced, and the pH of the solution is neutral.
$K_a> K_b$This is the case, for example, of $\ce{NH4F}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{HF}$. The ions $\ce{NH4^+}$ and $\ce{F^-}$ do hydrolyze and as ammonium ions hydrolyze to a greater extent than fluoride ions, i.e more hydronium ion is produced, and the pH of the solution is acidic.
$K_a< K_b$This is the case, for example, of $\ce{NH4CN}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{HCN}$. The ions $\ce{NH4^+}$ and $\ce{CN^-}$ do hydrolyze and as cyanide ions hydrolyze to a greater extent than fluoride ions, i.e less hydronium ions are produced, and the pH of the solution is basic.
$\ce{NaCl}$ is a salt of a strong acid and strong base. As you know, strong acids give weak conjugate bases and strong bases give weak conjugate acids. Hence, the $\ce{Cl-}$ ions are too weak to accept protons from water and $\ce{Na+}$ ions are to weak to react with $\ce{OH-}$. So there is no change in the concentration of $\ce{H+}$ or $\ce{OH-}$, so no change in pH.
But of a salt like $\ce{CH3COONa}$ is added, $\ce{CH3COO-}$ accepts protons and changes pH. This is because, weak acid $\ce{CH3COOH}$ gives strong conjugate base $\ce{CH3COO-}$.
$\ce{NaCl}$ is a salt of a strong acid and strong base. As you know, strong acids give weak conjugate bases and strong bases give weak conjugate acids. Hence, the $\ce{Cl-}$ ions are too weak to accept protons from water and $\ce{Na+}$ ions are to weak to react with $\ce{OH-}$. So there is no change in the concentration of $\ce{H+}$ or $\ce{OH-}$, so no change in pH.
But of a salt like $\ce{CH3COONa}$ is added, $\ce{CH3COO-}$ accepts protons and changes pH. This is because, weak acid $\ce{CH3COOH}$ gives strong conjugate base $\ce{CH3COO-}$.
When you dissolve a salt in water, you have to distinguish 4 cases:
The salt results from the reaction of a strong acid with a strong base, the pH will be 7. This is the case, for example, of $\ce{NaCl}$ that results from the reaction of $\ce{HCl}$ and $\ce{NaOH}$. The ions $\ce{Na^+}$ and $\ce{Cl^-}$ are spectator ions and don't react with water. So, the pH doesn't change.
The salt results from the reaction of a strong acid with a weak base, the pH will be less than 7. This is the case , for example, of $\ce{NH4Cl}$ that results from the reaction of $\ce{HCl}$ and $\ce{NH3}$. The ions $\ce{Cl^-}$ are spectator ions and don't react with water. While the ions $\ce{NH_4^+}$ do hydrolyze and give ions $\ce{H^+}$ to water. So, the pH will decrease.
The salt results from the reaction of a weak acid with a strong base, the pH will be more than 7. This is the case , for example, of $\ce{CH3COONa}$ that results from the reaction of $\ce{CH3COOH}$ and $\ce{NaOH}$. The ions $\ce{Na^+}$ are spectator ions and don't react with water. While the ions $\ce{CH3COO^-}$ do hydrolyze and accept ions $\ce{H^+}$ from water. In this case, water act as an acid leaving a hydroxide ion. So, the pH will increase.
The salt results from the reaction of a weak acid with a weak base, the calculus of pH is a little bit complex. We need to take into consideration the $K_a$ and the $K_b$ of the weak acid and the weak base and whichever is the stronger acid or weak will be the dominate factor in determining whether the solution is acidic or basic. We can investigate three cases:
When you dissolve a salt in water, you have to distinguish 4 cases:
The salt results from the reaction of a strong acid with a strong base, the pH will be 7. This is the case, for example, of $\ce{NaCl}$ that results from the reaction of $\ce{HCl}$ and $\ce{NaOH}$. The ions $\ce{Na^+}$ and $\ce{Cl^-}$ are spectator ions and don't react with water. So, the pH doesn't change.
The salt results from the reaction of a strong acid with a weak base, the pH will be less than 7. This is the case , for example, of $\ce{NH4Cl}$ that results from the reaction of $\ce{HCl}$ and $\ce{NH3}$. The ions $\ce{Cl^-}$ are spectator ions and don't react with water. While the ions $\ce{NH_4^+}$ do hydrolyze and give ions $\ce{H^+}$ to water. So, the pH will decrease.
The salt results from the reaction of a weak acid with a strong base, the pH will be more than 7. This is the case , for example, of $\ce{CH3COONa}$ that results from the reaction of $\ce{CH3COOH}$ and $\ce{NaOH}$. The ions $\ce{Na^+}$ are spectator ions and don't react with water. While the ions $\ce{CH3COO^-}$ do hydrolyze and accept ions $\ce{H^+}$ from water. In this case, water act as an acid leaving a hydroxide ion. So, the pH will increase.
The salt results from the reaction of a weak acid with a weak base, the calculus of pH is a little bit complex. We need to take into consideration the $K_a$ and the $K_b$ of the weak acid and the weak base and whichever is the stronger acid or weak will be the dominate factor in determining whether the solution is acidic or basic. We can investigate three cases:
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$\ce{NaCl}$ is a salt of a strong acid and strong base. As you know, strong acids give weak conjugate bases and strong bases give weak conjugate acids. Hence, the $\ce{Cl-}$ ions are too weak to accept protons from water and $\ce{Na+}$ ions are to weak to react with $\ce{OH-}$. So there is no change in the concentration of $\ce{H+}$ or $\ce{OH-}$, so no change in pH.
But of a salt like $\ce{CH3COONa}$ is added, $\ce{CH3COO-}$ accepts protons and changes pH. This is because, weak acid $\ce{CH3COOH}$ gives strong conjugate base $\ce{CH3COO-}$.
$\ce{NaCl}$ is a salt of a strong acid and strong base. As you know, strong acids give weak conjugate bases and strong bases give weak conjugate acids. Hence, the $\ce{Cl-}$ ions are too weak to accept protons from water and $\ce{Na+}$ ions are to weak to react with $\ce{OH-}$. So there is no change in the concentration of $\ce{H+}$ or $\ce{OH-}$, so no change in pH.
But of a salt like $\ce{CH3COONa}$ is added, $\ce{CH3COO-}$ accepts protons and changes pH. This is because, weak acid $\ce{CH3COOH}$ gives strong conjugate base $\ce{CH3COO-}$.
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