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Hint: Weak and strong bases are defined on their ability to dissociate when dissolved in water.
Complete step by step answer:
Base is a substance that releases hydroxide ion $\left( {O{H^ - }} \right)$ when dissolved in water. Oxides of alkaline earth metals are basic. Group 2 of periodic table contains alkaline earth metals. Bases are slippery in touch and taste bitter. They turn red litmus paper blue. Bases react with acid to form salt.
Weak bases do not dissociate completely when dissolved in water. That means there will be a small proportion of hydroxide ions and the concerned basic radical in the aqueous solution. Ionization of a weak base is a type of equilibrium process in which chemical equilibrium is established in solution examples of weak bases are ammonia $\left( {N{H_3}} \right)$ , pyridine \[\left( {{C_5}{H_5}N} \right)\] , trimethylamine $\left( {N{{\left( {C{H_3}} \right)}_3}} \right)$
Strong base dissociates completely when dissolved in water. It ionizes completely example of strong bases include hydroxides of alkali metals or alkaline earth metals that are sodium hydroxide $\left( {NaOH} \right)$ , potassium hydroxide $\left( {KOH} \right)$ , rubidium hydroxide $\left( {RbOH} \right)$ , strontium hydroxide $\left( {Sr{{\left( {OH} \right)}_2}} \right)$
Ions released by sodium hydroxide$ = N{a^ + } + O{H^ - }$
Ions released by potassium hydroxide$ = {K^ + } + O{H^ - }$
Ions released by rubidium hydroxide$ = R{b^ + } + O{H^ - }$
Ions released by strontium hydroxide$ = S{r^{ + 2}} + 2O{H^ - }$
Additional information: \[pH = - {\log _{10}}\left[ {{H^ + }} \right]\] since $pH$ is measure of hydrogen ion concentration, but base releases $O{H^ - }$ ions so, to find $pH$ of a base we need to know about ionic product of water ionic product of water is denoted as ${K_w}$. It is equal to \[{K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]\]
Value of ${K_w} = 1 \times {10^{ - 14}}mo{l^2}d{m^{ - 6}}$ at room temperature
Also, $p{K_w} = - {\log _{10}}{K_w}$
Note: $pH$ of base is above $7$ and below $14$ . For strong bases $pH$ is between $12$ and $14$
Complete step by step answer:
Base is a substance that releases hydroxide ion $\left( {O{H^ - }} \right)$ when dissolved in water. Oxides of alkaline earth metals are basic. Group 2 of periodic table contains alkaline earth metals. Bases are slippery in touch and taste bitter. They turn red litmus paper blue. Bases react with acid to form salt.
Weak bases do not dissociate completely when dissolved in water. That means there will be a small proportion of hydroxide ions and the concerned basic radical in the aqueous solution. Ionization of a weak base is a type of equilibrium process in which chemical equilibrium is established in solution examples of weak bases are ammonia $\left( {N{H_3}} \right)$ , pyridine \[\left( {{C_5}{H_5}N} \right)\] , trimethylamine $\left( {N{{\left( {C{H_3}} \right)}_3}} \right)$
Strong base dissociates completely when dissolved in water. It ionizes completely example of strong bases include hydroxides of alkali metals or alkaline earth metals that are sodium hydroxide $\left( {NaOH} \right)$ , potassium hydroxide $\left( {KOH} \right)$ , rubidium hydroxide $\left( {RbOH} \right)$ , strontium hydroxide $\left( {Sr{{\left( {OH} \right)}_2}} \right)$
Ions released by sodium hydroxide$ = N{a^ + } + O{H^ - }$
Ions released by potassium hydroxide$ = {K^ + } + O{H^ - }$
Ions released by rubidium hydroxide$ = R{b^ + } + O{H^ - }$
Ions released by strontium hydroxide$ = S{r^{ + 2}} + 2O{H^ - }$
Additional information: \[pH = - {\log _{10}}\left[ {{H^ + }} \right]\] since $pH$ is measure of hydrogen ion concentration, but base releases $O{H^ - }$ ions so, to find $pH$ of a base we need to know about ionic product of water ionic product of water is denoted as ${K_w}$. It is equal to \[{K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]\]
Value of ${K_w} = 1 \times {10^{ - 14}}mo{l^2}d{m^{ - 6}}$ at room temperature
Also, $p{K_w} = - {\log _{10}}{K_w}$
Note: $pH$ of base is above $7$ and below $14$ . For strong bases $pH$ is between $12$ and $14$
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