T1 = 200K ; k = k1
T2 = 400K ; k = k2
= 2k1

pH = -log10[H+]
∴[H+]=10-pH
Let the volume be x mL
V1M1 + V2M2
+ V3M3 =VM
∴
x mL of 10-1M+ x mL of
10-2M + x mL of 10-3 M
= 3x mL of [H+]
∴[H+ ] = x[0.1+0.01+0.001] / 3x
= ( 0.1+ 0.01+ 0.001 ) / 3
= 0..111 / 3
= 0.037
= 3.7 ×10−2
The ‘Z’ isCH3 - CH2 -
OH ---PCl5→ CH3
- CH2 - Cl --ale.KOH→CH2 = CH2 --H2SO4/H2O→ CH3 - CH2 – OH
(Z) ethanol