$NaHCO_3$ can be manufactured by Solvay's process but $K_2CO_3$ cannot be prepared because

  • A
    $K_2CO_3$ is more soluble
  • B
    $K_2CO_3$ is less soluble
  • C
    $KHCO_3$ is more soluble than $NaHCO_3$
  • D
    $KHCO_3$ is less soluble than $NaHCO_3$

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Statement $II$: Hydroxides of $Be$ and $Al$ dissolve in excess alkali to give beryllate and aluminate ions.
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