$2KOH+H_2SO_4\to K_2SO_4+2H_2O$
- Nếu $b\ge 2a$ (KOH dư hoặc hết)
$\Rightarrow n_{K_2SO_4}=n_{H_2SO_4}=a (mol)$
$\Rightarrow m_{\text{muối}}=174a (g)$
- Nếu $b<2a$ ($H_2SO_4$ dư)
$\Rightarrow n_{K_2SO_4}=0,5b (mol)$
$\Rightarrow m_{\text{muối}}=174.0,5b=87b (g)$