A buffer is made by dissolving #"NaClO"# and #"HClO"# in water. How do you write equations to show how this buffer neutralizes added H+ and OH-?
Examine the equilibrium of the buffer system,
wherein
Additionally, think about the ionization of water.
where
Understanding the effects of adding protons or hydroxide ions to the buffer solution can be achieved by referring to the equations above.
It might take some time to process what I'm going to tell you below; this is not something that is easily understood.
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To write equations showing how the buffer neutralizes added H+ and OH-, you would use the following reactions:
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Buffer neutralizing added H+: [ \text{HClO} + \text{H}^+ \rightarrow \text{ClO}^- + \text{H}_2\text{O} ]
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Buffer neutralizing added OH-: [ \text{NaClO} + \text{OH}^- \rightarrow \text{ClO}^- + \text{NaOH} ]
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The equations for the neutralization of added H⁺ and OH⁻ in a buffer made from NaClO and HClO can be represented as follows:
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Neutralization of added H⁺: [HClO + H^+ \rightarrow ClO^- + H_2O]
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Neutralization of added OH⁻: [NaClO + OH^- \rightarrow NaOH + ClO^-]
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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- Why is neutralization not a redox reaction?
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