College Algebra 7th Edition

$[\mathrm{H}^{+}]_{lime\ juice}=10^{-1.9}\approx 1.26\times 10^{-2}$ M
We are given: $\mathrm{p}\mathrm{H}_{lime\ juice}=1.9$ Therefore: $-\log_{10}[\mathrm{H}^{+}]=1.9$ $[\mathrm{H}^{+}]_{lime\ juice}=10^{-1.9}\approx 1.26\times 10^{-2}$ M