Following solutions were prepared by mixing different volumes of NAOH of HCL different concentrations.
(i) `60 mL (M)/(10) HCI+40 mL (M)/(10) NaOH`
(ii) `55 mL (M)/(10) HCI+45 mL (M)/(10) NaOH`
(iii) `75 mL (M)/(5) HCI+25 mL (M)/(5) NaOH`
(iv) `100 mL (M)/(10) HCI+100 mL (M)/(10) NaOH`
pH of which one of them will be equal to I?

Options

A. iv
B. (i)
C. (ii)
D. (iii)

Answer

D

Explanation

No of moles of HCl `=0.2 xx 75 xx 10^(-3) = 15 xx 10^(-3)`
No of mole of `NaOH = 0.2 xx 25 xx 10^(-3) = 5 xx 10^(-3)`
No of moles of HCl after mixing `=15 xx 10^(-3) -5 xx 10^(-3)`
`:.` Concentration of HCl `= ("No. of moles of HCl")/("Vol in litre") = (10 xx 10^(-3))/(100 xx 10^(-3)) = 0.1M`
For (iii) solutiion, pH of 0.1 M HCl = `-log_(10) (0.1) = 1`