What volume of a 0.1 M H\(_3\)PO\(_4\) will be required to neutralize 45.0 cm\(^3\) of a 0.2 M NaOH?
Firstly, write a balance chemical equation of the neutralization reaction
H\(_3\)PO\(_4\) + 3NaOH → Na\(_3\)PO\(_4\) + 3H\(_2\)O
Given:
V\(_A\) = ? C\(_A\) = 0.1M N\(_A\) = 1
V\(_B\) = 45cm\(^3\) C\(_B\) = 0.2M N\(_B\) = 3
Using the formula,
\(\frac{{C_A}{V_A}}{{C_B}{V_B}}\) = \(\frac{N_A}{N_B}\)
V\(_A\) = \(\frac{{C_B}{V_B}{N_A}}{{C_A}{N_B}}\)
V\(_A\) = \(\frac{{0.2}\times{45}\times{1}}{{0.1}\times{3}}\)
V\(_A\) = 30cm\(^3\)
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