what will happen to bromothymol blue indicator:
a.) after addition of HCl, how does it affects the equilibrium
b.) after addition of NaOH, does it affect the equilibrium
c.) after re-addition of NaOH
Rewrite the net equation for the equilibrium? is this an exothermic reactio or endothermic reaction.
the euqation is HInd=H^+ + Ind^- pls help
HIn ==> H^+ + In^-
Addition of HCl increases H^+ and shifts the equilibrium to the left.
Addition of NaOH decreases H^+ (due t5o H^+ + OH^- ==> H2O) and shifts the equilibrium to the right.
is this reaction endothermic or exothermic my answer is niether of the two because no heat is absorbed or released. ist it correct?
a.) When HCl is added to the bromothymol blue indicator, it will react with the indicator. The HCl will donate a proton (H+) to the indicator, causing it to shift towards the protonated form, resulting in a color change. The equilibrium will be driven to the right.
b.) When NaOH is added to the bromothymol blue indicator, it will react with the indicator. The NaOH will accept a proton (H+) from the indicator, causing it to shift towards the deprotonated form, resulting in a color change. The equilibrium will be driven to the left.
c.) If NaOH is re-added after the addition of HCl, it will react again with the protonated form of the indicator. This will cause the equilibrium to shift towards the deprotonated form and result in a color change.
The net equation for the equilibrium can be rewritten as: HInd ⇌ H+ + Ind^-
This reaction is an acid-base equilibrium. It is neither exothermic nor endothermic. The reaction only involves the transfer of protons (H+), not the exchange of energy.
To understand what will happen to bromothymol blue (BTB) indicator in each scenario, we need to consider the behavior of BTB in acidic and basic solutions.
a) After the addition of HCl (hydrochloric acid), the equilibrium of the indicator solution will shift towards the acidic side. This is because HCl is a strong acid that donates hydrogen ions (H+) when dissolved in water. The added H+ ions will react with the BTB indicator, causing it to change color. BTB changes from blue to yellow in acidic conditions.
b) After the addition of NaOH (sodium hydroxide), the equilibrium of the indicator solution will shift towards the basic side. NaOH is a strong base that dissociates into sodium ions (Na+) and hydroxide ions (OH-) in solution. The OH- ions will react with the BTB indicator, causing it to change color. BTB changes from blue to green in basic conditions.
c) If NaOH is re-added after HCl was initially added, the equilibrium will shift back towards the basic side, returning the BTB back to its green color.
The net equation for this equilibrium is as follows:
HInd ⇌ H+ + Ind^-
This equation shows the reversible reaction in which BTB (HInd) can dissociate into hydrogen ions (H+) and indicator ions (Ind^-). The equilibrium is established between the protonated form of BTB (HInd) and its deprotonated form (Ind^-).
As for the nature of the reaction, based on the given equation, we cannot determine if it is exothermic or endothermic. The equation represents the equilibrium between two forms of BTB and does not provide information about energy changes involved in the reaction.