MCQ
$pH + pOH$ equal to
- AZero
- ✓Fourteen
- CA negative number
- DInfinity
$pH =-\log _{10}\left[ H ^{+}\right]$
- $pOH$ is defined as the negative logarithm of Hydroxide ion $( OH -)$ concentration.
$pOH =-\log _{10}\left[ OH ^{-}\right]$
- The concentration of $\left[ H ^{+}\right]$and $\left[ OH ^{-}\right]$is $10^{-7}$ for a neutral solution.
$pH =-\log _{10}\left[10^{-7}\right]$
$pH =7 \log _{10}[10]$
$pH =7$
$pOH =-\log _{10}\left[10^{-7}\right]$
$pOH =7 \log _{10}[10]$
$pOH =7$
- The value of $\log _{10}[10]$ is $1 .$
- Calculate the sum of $pH$ and $pOH$.
$pH + pOH =7+7$
$=14$
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$C_2H_4(g) + H_2(g) \to C_2H_6(g)$
| Bond | Bond energy $(kJ)$ |
| $C-H$ | $413$ |
| $C-C$ | $348$ |
| $C=C$ | $610$ |
| $H-H$ | $436$ |

$(A)\,\,CH_3CH_2CH_2Br + KOH \rightarrow CH_3CH CH_2 + KBr + H_2O$