say for example the equilibrium I2(aq)+H2O(l)-----HOI(aq)+I(aq)+H(aq)
Think Lechatlier principle... addition of NaOH will cause the H ions to react with the OH ions to cause more water (more reactants) increase in reactants shifts the equilibrium in the FORWARD direction to form MORE H+ to restore the equilibrium
a base would decrease the ph or hydrogen concentration
A base or alkaline substance lowers the H (hydrogen ion) concentration in a solution. Bases can accept or remove hydrogen ions from the solution, increasing the concentration of OH- ions and thereby reducing the concentration of H+ ions. Examples of bases include sodium hydroxide (NaOH) and ammonia (NH3).
NaOH in pure state does not contain H+ ions but in solution state it does contain
The following equation will get you to the answer where [H+] is the concentration of hydrogen ion: pH = -log( [H+] ) or [H+] = 10-pH
concentration decreases
Use the working definition of pH used in General Chemistry classes: pH = -log([H+]) and the equilibrium constant for ionization of water: [H+][OH-]=10-14 (Here [] denotes concentration in Molarity) For moderate concentrations of NaOH (like 10-4-ish M and up, we can neglect the [OH-] from the actual ionization of water (since 10-7 is the maximum this concentration could be, when the NaOH concentration is 0, and even this is much less than the concentration of NaOH). Then we can say: 10-14=[H+][OH-]=[H+][NaOH] and then pH=-log[H+]=-log(10-14/[NaOH]) Just as an example, a 0.5 M solution has a pH of approximately -log(10-14/0.5) which is about 13
pH = - log([H+]) , pOH = - log([OH-] , pH + pOH = 14 [X] = concentration of X
a base would decrease the ph or hydrogen concentration
pH would decrease since H+ concentration increases, while pOH would increase since OH- concentration decreases. But pH would always be above 7 and pOH would always be below 7 since H+ will never be equal to OH- [OH-] would be always be greater than [H+] in any solution of NaOH , however dilute. Also remember pH + pOH =14, thus if pH decreases then pOH increases. Thanking you Yours Rajiv
[h+]=1 * 10-2[h+][oh-]=1 * 10-14[oh-]=1 * 10-12
A base or alkaline substance lowers the H (hydrogen ion) concentration in a solution. Bases can accept or remove hydrogen ions from the solution, increasing the concentration of OH- ions and thereby reducing the concentration of H+ ions. Examples of bases include sodium hydroxide (NaOH) and ammonia (NH3).
Hydrochloric acid is HCl. It is a H plus (H^+) donor, and so adding it to water will INCREASE the H^+ concentration. Increasing the H^+ concentration results in a DECREASE in pH.
2H - C - H + NaOH ------> H- C - O -Na + CH3 - OH
NaOH in pure state does not contain H+ ions but in solution state it does contain
No, there wouldn't be any measurable effect on the pH of the solution.The only way to affect the pH of the aqueous solution would be changing its proportion of OH- and "H+". At room temperature the concentration of these two species is 10^-7 mol/l and the product of both concentrations must always be 10^-14 mol²/l² at these conditions.Consequently a change in the concentration of one of these ions would affect the concentration of the other. This could be achieved by binding either OH- to Na+ or "H+" to Cl-. Binding OH- to Na+ would rise the concentration of "H+" and by this leading to an acidic character of the solution.Since NaOH is a very strong base and HCl is a very strong acid, both dissolve completely in water to Na+ (aq), OH-, Cl- (aq) and "H+". No chemically bound NaOH or HCl in the solution is observable and no change in the ("H+"/OH-)-ratio occurs. Thus the pH stays the same.
acids are substances that release their hydrogen ion(s) while bases grab hydrogen ions to themselves. SO, adding acids will increase the H+ concentration while adding bases will decrease the H+ concetration of the solution. This would be considered a direct effect.
MW NaOH is 23+16+1 = 40. 4.5g in 750ml is 6.0g in 1L. Molarity is 6/40 = 0.15 OH- concentration is 0.15, pOH is -log100.15 = 0.82 H+ concentration is 10-14/0.15 = 6.67x10-13 pH is -log10 6.67x10-13 = 12.18