Advertisements
Advertisements
प्रश्न
Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature?
Advertisements
उत्तर
Ionic product,
`"K"_"w" = ["H"^+] ["OH"^-]`
Let `["H"^+] = x`
Since `["H"^+] = ["OH"^(-)], "K"_"w" = x^2`
`=> "K"_"w" "at 310K" " is" 2.7 xx 10^(-14)`
`therefore 2.7 xx 10^(-14) = x^2`
`=> x = 1.64 xx 10^(-7)`
`=> ["H"^+] = 1.64 xx 10^(-7)`
`=> "pH" = -log["H"^+]`
`= - log [1.64 xx 10^(-7)]`
= 6.78
Hence, the pH of neutral water is 6.78.
संबंधित प्रश्न
The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2. in which of these solutions precipitation will take place?
What will be the correct order of vapour pressure of water, acetone and ether at 30°C. Given that among these compounds, water has maximum boiling point and ether has minimum boiling point?
Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution. \[\ce{(K_{sp} of PbCl2}\] = 3.2 × 10–8, atomic mass of \[\ce{Pb = 207 u}\]).
