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  3. Chemistry
  4. At t°C, Kw for water is 6.4 × 10-13. The pH of water at t°C will be

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Q. At $t^{\circ}C$, $K_w$ for water is $6.4 \times 10^{-13}$. The $pH$ of water at $t^{\circ}C$ will be

Solution:

$PK_{w}=-logK_{w}=-log(6.4\times10^{-13}=12.194$
$PH=POH$ for neutral solution
$PH=\frac{12.194}{2}=6.097$