40 ml of `0.1` M ammonia is mixed with 20 ml of `0.1 M HCI`. What is the pH of the mixture ? (`pK_(b)` of ammonia solution is `4.74.`)

Question

40 ml of `0.1` M ammonia is mixed with 20 ml of `0.1 M HCI`. What is the pH of the mixture ? (`pK_(b)` of ammonia solution is `4.74.`)

40 ml of `0.1` M ammonia is mixed with 20 ml of `0.1 M HCI`. What is the pH of the mixture ? (`pK_(b)` of ammonia solution is `4.74.`)
A. 4.74
B. 2.26
C. 9.26
D. `5.00`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

Related Questions

29.2% (W/W) HCI stock solution has a density of 1.25 g `mL^(-1)`. The molecular mass of HCI is 36.5 g `mol^(-1)` Calculate the volume (mL) of stock so

2 Answers 4 Views

25mL of `3M HCI` were added to 75 mL of `0.05M HCI`. The molarity of HCI in the resulting solution is approximately

2 Answers 5 Views

`pK_(a1)` and `pK_(a_(2))` of `H_(2)CO_(3)` are 6.38 and 10.26 respectively. The pH of 1 M and 0.1 M `NaHCO_(3)` are

2 Answers 4 Views

`pK_(a1) , pK_(a_(2))` and `pK_(a_(3))` of `H_(3)PO_(4)` are respectively x,y and z. pH of 0.1 M `Na_(2)HPO_(4)` solution is

2 Answers 5 Views

The `pK_(a)` of an acid HA is 4.77 and `pK_(b)` of a base of BOH is 4.75 . The pH of 0.1 M aqueous solution of the salt AB is

2 Answers 7 Views

A solution of HCI is prepared by dissolving 5.5 g HCI in 200 g ehanol. The density of the solution is 0.79 g `mL^(-1)` Molarity of solution is :

2 Answers 5 Views

The molarity of a solution obtained by mixing 750 mL of 0.5dsM solution of HCI with 250 mL of 2M solution of HCI will be :

2 Answers 8 Views

25mL of 3.0 M HCI are mixed with 75 of 4.0 M HCI. If the volumes are volumes are aditive, the molarity of the final solution will be :

2 Answers 5 Views

you are supplied with 500 mL each so 2N HCI and 5N HCI. What is the maximym volume of 3M HCI that you can can prpare using only these two solutions ?

2 Answers 4 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

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

Correct Answer - C
40 Ml of 0.1 M HCl `=20 xx 0.1` milli mole `=2` milli mole
`NH_(4)OH+HCl rarr NH_(4)Cl +H_(2)O`
2 milli mole of HCl will neutralize 2 milli moles of `NH_(4)OH` to form 2 milli moles of `NH_(4)Cl`.
`NH_(4)OH ` left `=2 ` milli moles
Total volume `=60 mL`
`:. [NH_(4)OH]=2//60M,[NH_(4)Cl]=2//60M`
` pOH =pK_(b)+log.([NH_(4)Cl])/([NH_(4)OH])`
`=4.74 +log. (2//60)/(2//60)=4.74`
`:. pH =14-4.74 =9.26`