Answer
$[\mathrm{H}^{+}]_{lime\ juice}=10^{-1.9}\approx 1.26\times 10^{-2}$ M
Work Step by Step
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