If 29.60 mL of 0.0985 M NaOH is required to neutralize 0.320 g of an unknown acid, HA, what is the molecular weight of the unknown acid?
Added by Jesus A.
Step 1
Step 1: Write the balanced chemical equation for the neutralization reaction between NaOH and HA: NaOH + HA -> NaA + H2O where NaA is the sodium salt of the unknown acid HA. Show more…
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You have 0.6721 g of an unknown monoprotic acid, HA, which reacts with NaOH according to the balanced equation HA + NaOH → NaA + H2O If 36.68 mL of 0.0985 M NaOH is required to titrate the acid to the equivalence point, what is the molar mass of the acid?
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A mass of 0.4113 g of an unknown acid, HA, is titrated with NaOH(aq). If the acid reacts with 28.10 mL of 0.1055 M NaOH(aq), what is the molar mass of the acid?
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You have $0.954 \mathrm{g}$ of an unknown acid, $\mathrm{H}_{2} \mathrm{A},$ which reacts with NaOH according to the balanced equation $\mathrm{H}_{2} \mathrm{A}(\mathrm{aq})+2 \mathrm{NaOH}(\mathrm{aq}) \longrightarrow \mathrm{Na}_{2} \mathrm{A}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell)$ If 36.04 mI. of $0.509 \mathrm{M} \mathrm{NaOH}$ is required to titrate the acid to the equivalence point, what is the molar mass of the acid?
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