QUESTION IMAGE
Question
complete the reactions for dissolving the following two acids into water. then, label the acid, base, conjugate acid, and conjugate base.
- $\ce{h3po3 + h2o \leftrightarrow h3o^{+1} + \\_\\_1\\_\\_}$
acid \\_\\_2\\_\\_
base \\_\\_3\\_\\_
conj acid \\_\\_4\\_\\_
conj base \\_\\_5\\_\\_
- $\ce{hco3^{-1} + h2o \leftrightarrow h3o^{+1} + \\_\\_6\\_\\_}$
acid \\_\\_7\\_\\_
base \\_\\_8\\_\\_
conj acid \\_\\_9\\_\\_
conj base \\_\\_10\\_
a. $\ce{h3po4}$ b. $\ce{hcl}$ c. phosphoric d. perchloric e. hydrofluoric
f. $\ce{h2so4}$ g. $\ce{h3po3}$ h. $\ce{h2o}$ i. $\ce{oh^{-1}}$ j. $\ce{h3o^{+1}}$ k. $\ce{so4^{-2}}$
l. $\ce{mg^{+2}}$ m. $\ce{cl^{-1}}$ n. $\ce{hpo3^{-2}}$ o. $\ce{h2po3^{-1}}$ p. $\ce{co3^{-2}}$ q. $\ce{hco3^{-1}}$
r. $\ce{h2s}$ s. $\ce{hs^{-1}}$ t. $\ce{s^{-2}}$ u. 0.2500 v. 0.1000 w. 0.02500
x. $1.000 \times 10^{-3}$ y. 3.0000 z. $1 \times 10^{-14}$ aa. 2.6990 bb. $1.000 \times 10^{-11}$
cc. 11.0000 dd. 2 ee. 1 ff. $2.000 \times 10^{-3}$ gg. $5.000 \times 10^{-12}$
hh. 11.3010 ii. 0.01250 jj. 0.22100 kk. 0.05656 ll. acid
mm. conjugate acid oo. base pp. conjugate base
Step1: Analyze the first reaction
According to the principle of acid - base reaction (proton transfer), \(H_3PO_3\) donates a proton (\(H^+\)) to \(H_2O\). The formula of the conjugate base of \(H_3PO_3\) is \(H_2PO_3^{-1}\) (subtract one \(H^+\) from \(H_3PO_3\)).
Step2: Label the species in the first reaction
- Acid: \(H_3PO_3\) (donates a proton)
- Base: \(H_2O\) (accepts a proton)
- Conjugate acid: \(H_3O^{+1}\) (\(H_2O\) accepts a proton to form it)
- Conjugate base: \(H_2PO_3^{-1}\) (\(H_3PO_3\) donates a proton to form it)
Step3: Analyze the second reaction
\(HCO_3^{-1}\) donates a proton (\(H^+\)) to \(H_2O\). The formula of the conjugate base of \(HCO_3^{-1}\) is \(CO_3^{-2}\) (subtract one \(H^+\) from \(HCO_3^{-1}\)).
Step4: Label the species in the second reaction
- Acid: \(HCO_3^{-1}\) (donates a proton)
- Base: \(H_2O\) (accepts a proton)
- Conjugate acid: \(H_3O^{+1}\) (\(H_2O\) accepts a proton to form it)
- Conjugate base: \(CO_3^{-2}\) (\(HCO_3^{-1}\) donates a proton to form it)
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- \(O\)
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