Home
Class 11
CHEMISTRY
Salt of weak acids and strong bases hyd...

Salt of weak acids and strong bases hydrolysis in water to form …………… solutions.

Text Solution

AI Generated Solution

To solve the question "Salt of weak acids and strong bases hydrolysis in water to form …………… solutions," we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Components**: - We need to consider a salt formed from a weak acid and a strong base. - A common example is sodium acetate (CH₃COONa), which is formed from acetic acid (CH₃COOH, a weak acid) and sodium hydroxide (NaOH, a strong base). ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • EQUILIBRIUM

    ICSE|Exercise NUMERICAL PROBLEMS |13 Videos

  • EQUILIBRIUM

    ICSE|Exercise NCERT TEXT-BOOK EXERCISE |109 Videos

  • EQUILIBRIUM

    ICSE|Exercise TRUE OR FALSE TYPE QUESTION |22 Videos

  • ENVIRONMENTAL CHEMISTRY

    ICSE|Exercise NCERT TEXT-BOOK EXERCISE|20 Videos

  • HYDROCARBONS

    ICSE|Exercise NCERT TEXT-BOOK EXERCISES |46 Videos

Similar Questions

Explore conceptually related problems

0.1 M solution of a salt of a weak acid a strong base hydrolysis. If K_(a) of the weak acid is 1 xx 10^(-5) , the percent hydrolysis of the salt is

When a salt of weak acid and weak base is dissolved in water, the pH of the resulting solution will be : (a)be 7 (b)be greater than 7 (c)be less than 7 (d)depend upon K_(a) and K_(b) values

Match item (A) NH_(4)Cl , (B) NH_(4)OH , ( C) NaCl, (D) FeCl_(3) , (E) NaHSO_(4) , (F) NaOH with the correct description given in (i)-(vi) below. (i) It is a weak base. (ii) It is a salt of strong acid and a strong base. (iii) It is formed by direct combination. (iv) Its aqueous solution has acidic nature. (v) It gives pink colour with phenolphthalein. (vi) It is an acid salt.

When a salt reacts with water to form acidic or basic solution , the process is called hydrolysis . The pH of salt solution can be calculated using the following relations : pH = 1/2 [pK_(w) +pK_(a) + logc] (for salt of weak acid and strong base .) pH = 1/2 [pK_(w) - pK_(b) - logc] (for salt of weak base and strong acid ) . pH = 1/2 [ pK_(w)+pK_(a)-pK_(b)] (for weak acid and weak base ). where 'c' represents the concentration of salt . When a weak acid or a weak base not completely neutralized by strong base or strong acid respectively , then formation of buffer takes place . The pH of buffer solution can be calculated using the following relation : pH = pK_(a) + log . (["Salt"])/(["Acid"]) , pOH = pK_(b) + log . (["Salt"])/([ "Base"]) Answer the following questions using the following data : pK_(a) = 4.7447 , pK_(b) = 4.7447 ,pK_(w) = 14 When 100 mL of 0.1 M NH_(4)OH is added to 50 mL of 0.1M HCl solution , the pH is

When a salt reacts with water to form acidic or basic solution , the process is called hydrolysis . The pH of salt solution can be calculated using the following relations : pH = 1/2 [pK_(w) +pK_(a) + logc] (for salt of weak acid and strong base .) pH = 1/2 [pK_(w) - pK_(b) - logc] (for salt of weak base and strong acid ) . pH = 1/2 [ pK_(w)+pK_(a)-pK_(b)] (for weak acid and weak base ). where 'c' represents the concentration of salt . When a weak acid or a weak base not completely neutralized by strong base or strong acid respectively , then formation of buffer takes place . The pH of buffer solution can be calculated using the following relation : pH = pK_(a) + log . (["Salt"])/(["Acid"]) , pOH = pK_(b) + log . (["Salt"])/([ "Base"]) Answer the following questions using the following data : pK_(a) = 4.7447 , pK_(b) = 4.7447 ,pK_(w) = 14 0.001 M NH_(4)Cl aqueous solution has pH :

When a salt reacts with water to form acidic or basic solution , the process is called hydrolysis . The pH of salt solution can be calculated using the following relations : pH = 1/2 [pK_(w) +pK_(a) + logc] (for salt of weak acid and strong base .) pH = 1/2 [pK_(w) - pK_(b) - logc] (for salt of weak base and strong acid ) . pH = 1/2 [ pK_(w)+pK_(a)-pK_(b)] (for weak acid and weak base ). where 'c' represents the concentration of salt . When a weak acid or a weak base not completely neutralized by strong base or strong acid respectively , then formation of buffer takes place . The pH of buffer solution can be calculated using the following relation : pH = pK_(a) + log . (["Salt"])/(["Acid"]) , pOH = pK_(b) + log . (["Salt"])/([ "Base"]) Answer the following questions using the following data : pK_(a) = 4.7447 , pK_(b) = 4.7447 ,pK_(w) = 14 50 mL 0.1 M NaOH is added to 50 mL of 0.1 M CH_(3)COOH solution , the pH will be

A weak acid HA has K_(a) = 10^(-6) . What would be the molar ratio of this acid and its salt with strong base so that pH of the buffer solution is 5 ?

Calcium lactate is a salt of weak organic acid and strong base represented as Ca(LaC)_(2). A saturated solution of Ca(LaC)_(2) contains 0.6 mole in 2 litre solution. pOH of solution is 5.60. If 90% dissociation of the salt takes place then what is pK_(a) of lactic acid?

Enthalpy of neutralzation is defined as the enthalpy change when 1 mole of acid /base is completely neutralized by base // acid in dilute solution . For Strong acid and strong base neutralization net chemical change is H^(+) (aq)+OH^(-)(aq)to H_(2)O(l) Delta_(r)H^(@)=-55.84KJ//mol DeltaH_("ionization")^(@) of aqueous solution of strong acid and strong base is zero . when a dilute solution of weak acid or base is neutralized, the enthalpy of neutralization is somewhat less because of the absorption of heat in the ionzation of the because of the absorotion of heat in the ionization of the weak acid or base ,for weak acid /base DeltaH_("neutrlzation")^(@)=DeltaH_("ionization")^(@)+ Delta _(r)H^(@)(H^(+)+OH^(-)to H_(2)O) What is DeltaH^(@) for complate neutralization of strong diacidic base A(OH)_(2)by HNO_(3) ?

Anions of a weak acid react with water to produce ………….. solution .

Home
Profile