`50mL` of `0.05M Na_(2)CO_(3)` is titrated against `0.1 M HCl` ,`pH` of the solution will be [Given :For `H_(2)CO_(3)` ,`pK_(a)=6.35,pK_(a)=10.33]`

Question

`50mL` of `0.05M Na_(2)CO_(3)` is titrated against `0.1 M HCl` ,`pH` of the solution will be [Given :For `H_(2)CO_(3)` ,`pK_(a)=6.35,pK_(a)=10.33]`

`50mL` of `0.05M Na_(2)CO_(3)` is titrated against `0.1 M HCl` ,`pH` of the solution will be [Given :For `H_(2)CO_(3)` ,`pK_(a)=6.35,pK_(a)=10.33]`
A. `6.35`
B. `6.526`
C. `8.34`
D. `6.173`

Monjur Alahi

8 views

2025-01-04 04:42

1 Answers

Related Questions

Assertion : Oxidation of `Na_(2)SO_(3)` is not caused by air but in the presence of `Na_(3)AsO_(3).` Reason : `Na_(3)AsO_(3)` can be oxidised by air.

2 Answers 6 Views

25.2 gm of a mixture of `NaHCO_(3)` and `Na_(2)CO_(3)` is heated strongly, 0.66 gm of `CO_(2)` gas is evolved then the % mass of `Na_(2) CO_(3)` prese

2 Answers 6 Views

A certain mixture of HCl and `CH_3-COOH` is `0.1 M` in each of the acids.20 ml of this solution is titrated against 0.1M NaOH.By how many units does t

2 Answers 6 Views

The `pK_(a_1), pK_(a_2) and pK_(a_3)` value for the amino acid cysteine `(HS-CH_2-undersetunderset(NH_2)(|)CH-COOH)` are respectively 1.8, 8.3, 10.8.

2 Answers 8 Views

Upon passing `0.01` moles `HCl` gas through `100mL` of `0.05M` Formic acid solution `(K_(a)=1.8xx10^(-4))` determine change in `pH` of solution and `[

2 Answers 16 Views

`50mL` of `0.05M` Propane `-1,2-`diamine solution is titrated with `0.1M HCl` solution at `25^(@)C`. Detemine the `pH` of solution obtained by adding

2 Answers 20 Views

The `pH` of a solution resulting from the addition of `12.5mL` of `0.1M HCl` to `50mL` of a solution containing 0.15M `CH_(3)COOH` & `0.2M` `CH_(3)COO

2 Answers 10 Views

Upon titrating `50mL` of `1.96%(w//v) H_(2)SO_(4)` with a `KOH` solution (containing `11.2g KOH` per litre of solution on adding `50mL KOH` solution.

2 Answers 8 Views

Which of the following solutions have different `pH`? `100mL" of "0.2m" "HCl +100mL" of "0.4 M" "NH_(3)` `50mL" of "0.1 M" HCl +"50mL" of "03.2 M NH(3

2 Answers 5 Views

Aniline behaves as a weak base. When `0.1M ,50mL` solution sample of aniline was mixed with `0.2 M ,12.5mL` of solution of `HCl` ,the `pH` of resultin

2 Answers 10 Views

Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

`{:(,CO_(3)^(2-),+,H^(+),rarr,HCO_(3)^(-),),("Initial milli-moles",50xx0.05,,40xx0.1,,-,),("Final milli-moles",-,,1.5,,2.5,),(,HCO_(3)^(-),+,H^(+),rarr,H_(2)CO_(3),),("Initial milli-moles",2.5,,1.5,,-,),("Final milli-moles",-,,1,,1.5,):}`
` pH=pK_(a_(1))+log(([HCO_(3)^(-)])/([H_(2)CO_(3)]))=6.173`

8 views Answered 2023-02-05 15:29